New Discovering Mathematics 3A Chapter 1: More About Quadratic Equations and Quadratic Functions — Solutions
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Try it yourself 1–24
Try it yourself 1
(a)
\begin{align} 2x^2 + 9x - 5 & = 0 \\ (2x - 1)(x + 5) & = 0 \end{align} \begin{align} 2x - 1 & = 0 && \text{ or } & x + 5 & = 0 \\ 2x & = 1 &&& x & = -5 \\ x & = {1 \over 2} \end{align}
(b)
\begin{align} 2x^2 - 3x + 5 & = 4 \\ 2x^2 - 3x + 5 - 4 & = 0 \\ 2x^2 - 3x + 1 & = 0 \\ (2x - 1)(x - 1) & = 0 \end{align} \begin{align} 2x - 1 & = 0 && \text{ or } & x - 1 & = 0 \\ 2x & = 1 &&& x & = 1 \\ x & = {1 \over 2} \end{align}
(c)
\begin{align} (x - 6)(x + 2) & = 9 \\ x^2 + 2x - 6x - 12 & = 9 \\ x^2 - 4x - 12 & = 9 \\ x^2 - 4x - 12 - 9 & = 0 \\ x^2 - 4x - 21 & = 0 \\ (x + 3)(x - 7) & = 0 \end{align} \begin{align} x + 3 & = 0 && \text{ or } & x - 7 & = 0 \\ x & = -3 &&& x & = 7 \end{align}
Try it yourself 2
(a)
\begin{align} x^2 - 12x - 5 & = x^2 - 12x + \left(12 \over 2\right)^2 - \left(12 \over 2\right)^2 - 5 \\ & = x^2 - 12x + 6^2 - 6^2 - 5 \\ & = (x - 6)^2 - 36 - 5 \\ & = (x - 6)^2 - 41 \end{align}
(b)
\begin{align} x^2 + 3x + 1 & = x^2 + 3x + \left(3 \over 2\right)^2 - \left(3 \over 2\right)^2 + 1 \\ & = \left(x + {3 \over 2}\right)^2 - {9 \over 4} + 1 \\ & = \left(x + {3 \over 2}\right)^2 - {5 \over 4} \end{align}
Try it yourself 3
(a)
\begin{align} -x^2 + 6x + 5 & = - (x^2 - 6x) + 5 \\ & = - \left[ x^2 - 6x + \left(6 \over 2\right)^2 - \left(6 \over 2\right)^2 \right] + 5 \\ & = - [ x^2 - 6x + 3^2 - 3^2 ] + 5 \\ & = - [ (x - 3)^2 - 9 ] + 5 \\ & = - (x - 3)^2 + 9 + 5 \\ & = - (x - 3)^2 + 14 \end{align}
(b)
\begin{align} -x^2 - x - 1 & = - (x^2 + x) - 1 \\ & = - \left[ x^2 + x + \left(1 \over 2\right)^2 - \left(1 \over 2\right)^2 \right] - 1 \\ & = - \left[ \left(x + {1 \over 2}\right)^2 - {1 \over 4} \right] - 1 \\ & = - \left(x + {1 \over 2}\right)^2 + {1 \over 4} - 1 \\ & = - \left(x + {1 \over 2}\right)^2 - {3 \over 4} \end{align}
Try it yourself 4
(a)
\begin{align} x^2 + 2x - 15 & = x^2 + 2x + \left(2 \over 2\right)^2 - \left(2 \over 2\right)^2 - 15 \\ & = x^2 + 2x + 1^2 - 1^2 - 15 \\ & = (x + 1)^2 - 1 - 15 \\ & = (x + 1)^2 - 16 \end{align}
(b)
\begin{align} x^2 + 2x - 15 & = 0 \\ \underbrace{(x + 1)^2 - 16}_\text{Use result from (a)} & = 0 \\ (x + 1)^2 & = 16 \\ x + 1 & = \pm \sqrt{16} \\ x + 1 & = \pm 4 \end{align} \begin{align} x + 1 & = 4 && \text{ or } & x + 1 & = -4 \\ x & = 4 - 1 &&& x & = -4 - 1 \\ x & = 3 &&& x & = -5 \end{align}
Try it yourself 5
(a)
\begin{align} x^2 + x - 4 & = x^2 + x + \left(1 \over 2\right)^2 - \left(1 \over 2\right)^2 - 4 \\ & = \left(x + {1 \over 2}\right)^2 - {1 \over 4} - 4 \\ & = \left(x + {1 \over 2}\right)^2 - {17 \over 4} \end{align}
(b)
\begin{align} x^2 + x - 4 & = 0 \\ \underbrace{ \left(x + {1 \over 2}\right)^2 - {17 \over 4}}_\text{From (a)} & = 0 \\ \left(x + {1 \over 2}\right)^2 & = {17 \over 4} \\ x + {1 \over 2} & = \pm \sqrt{17 \over 4} \end{align} \begin{align} x + {1 \over 2} & = \sqrt{17 \over 4} && \text{ or } & x + {1 \over 2} & = - \sqrt{17 \over 4} \\ x & = \sqrt{17 \over 4} - {1 \over 2} &&& x & = - \sqrt{17 \over 4} - {1 \over 2} \\ x & \approx 1.56 &&& x & \approx -2.56 \end{align}
Try it yourself 6
\begin{align} x^2 + x + 1 & = 0 \\ x^2 + x + \left(1 \over 2\right)^2 - \left(1 \over 2\right)^2 + 1 & = 0 \\ \left(x + {1 \over 2}\right)^2 - {1 \over 4} + 1 & = 0 \phantom{000000} [\text{Complete the square}] \\ \left(x + {1 \over 2}\right)^2 + {3 \over 4} & = 0 \\ \left(x + {1 \over 2}\right)^2 & = - {3 \over 4} \\ x + {1 \over 2} & = \pm \sqrt{-{3 \over 4}} \\ \\ \\ \text{No real roots for } & \sqrt{-{3 \over 4}} \\ \\ \therefore \text{Equation has no} & \text{ real roots} \end{align}
Try it yourself 7
\begin{align} 4x^2 & + 7x - 3 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-7 \pm \sqrt{ (7)^2 - 4(4)(-3)} \over 2(4) } \\ & = {-7 \pm \sqrt{97} \over 8} \\ & = 0.3561 \text{ or } -2.1061 \\ & \approx 0.36 \text{ or } -2.11 \text{ (2 d.p.)} \end{align}
Try it yourself 8
\begin{align} (9x + 5)(x + 2) & = -x - 6 \\ 9x^2 + 18x + 5x + 10 & = -x - 6 \\ 9x^2 + 23x + 10 & = - x - 6 \\ 9x^2 + 23x + x + 10 + 6 & = 0 \\ 9x^2 + 24x + 16 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-24 \pm \sqrt{(24)^2 - 4(9)(16)} \over 2(9)} \\ & = {-24 \pm \sqrt{0} \over 18} \\ & = -{4 \over 3} \end{align}
Try it yourself 9
\begin{align} 3x^2 + 11 & = 5x \\ 3x^2 - 5x + 11 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-5) \pm \sqrt{(-5)^2 - 4(3)(11)} \over 2(3)} \\ & = {5 \pm \sqrt{-107} \over 6} \\ \\ \\ \sqrt{-107} \text{ is } & \text{not a real number} \\ \\ \therefore \text{Equation } & \text{has no real roots} \end{align}
Try it yourself 10
(a)
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|
| y | 2 | -2 | -4 | -4 | -2 | 2 | 8 |
(b)
(c) Look for the x-intercepts of the curve
$$ x = -2.55, 1.55 $$
Try it yourself 11
(a)
| x | -5 | -4 | -3 | -2 | -1 | 0 | 1 |
|---|---|---|---|---|---|---|---|
| y | 9 | 4 | 1 | 0 | 1 | 4 | 9 |
(b)
(c) Look for the x-intercept of the curve
$$ x = - 2 $$
Try it yourself 12
(a)
| x | -2 | -1 | -0.5 | 0 | 0.5 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|---|
| y | -19 | -8 | -4.75 | -3 | -2.75 | -4 | -11 | -24 |
(b)(i)
\begin{align} \underbrace{ -3x^2 + 2x - 3}_\text{Curve} & = -12 \\ \\ \text{Draw } & y = -12 \\ \\ \\ \text{From graph, } x & = -1.4, 2.1 \end{align}
(b)(ii)
\begin{align} \underbrace{ -3x^2 + 2x - 3}_\text{Curve} & = x - 6 \\ \\ \text{Draw } & y = x - 6 \end{align}
| x | 0 | 1 | 2 |
|---|---|---|---|
| y | -6 | -5 | -4 |
$$ \text{From graph, } x = -0.85, 1.2 $$
Try it yourself 13
(a)
\begin{align} y & = 3x^2 + x - 2 \phantom{00} \text{--- (1)} \\ \\ y & = 3x + 3 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 3x^2 + x - 2 & = 3x + 3 \\ 3x^2 + x - 3x - 2 - 3 & = 0 \\ 3x^2 - 2x - 5 & = 0 \\ \\ \therefore b = -2, c & = - 5 \end{align}
(b)
\begin{align} \text{From graph, mini} & \text{mum point is } (-0.2, -2.1) \\ \\ \therefore k & < -2.1 \end{align}
(c)(i)
\begin{align} 3x^2 + 3x - 14 & = 0 \\ 3x^2 + 3x - 14 - 2x & = -2x \\ 3x^2 + x - 14 & = -2x \\ 3x^2 + x - 14 + 12 & = -2x + 12 \\ \underbrace{3x^2 + x - 2}_\text{Curve} & = - 2x + 12 \\ \\ \text{Draw } & y = -2x + 12 \end{align}
| x | 0 | 1 | 2 |
|---|---|---|---|
| y | 12 | 10 | 8 |
$$ \text{From graph, } x = -2.7, 1.7 $$
(c)(ii)
\begin{align} 3x^2 + 4x - 5 & = 0 \\ 3x^2 + 4x - 5 - 3x & = -3x \\ 3x^2 + x - 5 & = - 3x \\ 3x^2 + x - 5 + 3 & = - 3x + 3 \\ \underbrace{3x^2 + x - 2}_\text{Curve} & = -3x + 3 \\ \\ \text{Draw } & y = -3x + 3 \end{align}
| x | 0 | 1 | 2 |
|---|---|---|---|
| y | 3 | 0 | -3 |
$$ \text{From graph, } x = -2.1, 0.8 $$
Try it yourself 14
\begin{align}
{3x - 1 \over 2x + 5} & = {x - 3 \over x + 4} \\
{(3x - 1)(x + 4) \over (2x + 5)(x + 4)} & = {(x - 3)(2x + 5) \over (2x + 5)(x + 4)} \\
\\
(3x - 1)(x + 4) & = (x - 3)(2x + 5) \\
3x^2 + 12x - x - 4 & = 2x^2 + 5x - 6x - 15 \\
3x^2 + 11x - 4 & = 2x^2 - x - 15 \\
3x^2 - 2x^2 + 11x + x - 4 + 15 & = 0 \\
x^2 + 12x + 11 & = 0 \\
(x + 1)(x + 11) & = 0
\end{align}
\begin{align}
x + 1 & = 0
&& \text{ or } &
x + 11 & = 0 \\
x & = -1
&&&
x & = -11
\end{align}
$$ \text{When } x = -1 \text{ or } x = -11, \phantom{0} 2x + 5 \neq 0 \text{ and } x + 4 \neq 0 $$
$$ \therefore x = -1 \text{ or } x = -11 $$
Try it yourself 15
\begin{align} {x + 2 \over x} + {2 \over x - 2} & = {2 \over 1} \\ {(x + 2)(x - 2) \over x(x - 2)} + {2x \over x(x - 2)} & = {2x(x - 2) \over x(x - 2)} \\ \\ (x + 2)(x - 2) + 2x & = 2x(x - 2) \\ \underbrace{x^2 - 2^2}_{(a + b)(a - b) = a^2 - b^2} + 2x & = 2x^2 - 4x \\ x^2 - 4 + 2x & = 2x^2 - 4x \\ 0 & = 2x^2 - x^2 - 4x - 2x + 4 \\ 0 & = x^2 - 6x + 4 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-6) \pm \sqrt{(-6)^2 - 4(1)(4)} \over 2(1)} \\ & = {6 \pm \sqrt{20} \over 2} \\ & = 5.236 \text{ or } 0.76393 \\ & \approx 5.24 \text{ or } 0.764 \end{align}
Try it yourself 16
(a)
\begin{align} {5 \over x + 3} + {3x + 11 \over x^2 - 9} & = {5 \over x + 3} + {3x + 11 \over x^2 - 3^2} \\ & = {5 \over x + 3} + {3x + 11 \over (x + 3)(x - 3)} \phantom{000000} [ a^2 - b^2 = (a + b)(a - b)] \\ & = {5(x - 3) \over (x + 3)(x - 3)} + {3x + 11 \over (x + 3)(x - 3)} \\ & = {5(x - 3) + 3x + 11 \over (x + 3)(x - 3)} \\ & = {5x - 15 + 3x + 11 \over (x + 3)(x - 3)} \\ & = {8x - 4 \over (x + 3)(x - 3)} \end{align}
(b)
\begin{align} {5 \over x + 3} + {3x + 11 \over x^2 - 9} & = 3 \\ {8x - 4 \over (x + 3)(x - 3)} & = {3 \over 1} \\ {8x - 4 \over (x + 3)(x - 3)} & = {3(x + 3)(x - 3) \over (x + 3)(x - 3)} \\ \\ 8x - 4 & = 3(x + 3)(x - 3) \\ 8x - 4 & = 3(x^2 - 9) \\ 8x - 4 & = 3x^2 - 27 \\ 0 & = 3x^2 - 8x - 27 + 4 \\ 0 & = 3x^2 - 8x - 23 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-8) \pm \sqrt{(-8)^2 - 4(3)(-23)} \over 2(3)} \\ & = {8 \pm \sqrt{ 340 } \over 6 } \\ & = 4.4065 \text{ or } -1.7398 \\ & \approx 4.41 \text{ or } -1.74 \end{align}
Try it yourself 17
\begin{align} {x \over x - 2} - {1 \over x + 5} & = {5 \over (x - 2)(x + 5)} \\ {x(x + 5) \over (x - 2)(x + 5)} - {x - 2 \over (x - 2)(x + 5)} & = {5 \over (x - 2)(x + 5)} \\ {x(x + 5) - (x - 2) \over (x - 2)(x + 5)} & = {5 \over (x - 2)(x + 5)} \\ \\ x(x + 5) - (x - 2) & = 5 \\ x^2 + 5x - x + 2 & = 5 \\ x^2 + 4x + 2 & = 5 \\ x^2 + 4x + 2 - 5 & = 0 \\ x^2 + 4x - 3 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-4 \pm \sqrt{(4)^2 - 4(1)(-3)} \over 2(1)} \\ & = {-4 \pm \sqrt{28} \over 2} \\ & = 0.64575 \text{ or } -4.6458 \\ & \approx 0.646 \text{ or } -4.65 \end{align}
Try it yourself 18
\begin{align} \text{Let } x \text{ represent the } & \text{width of the path} \\ \\ \text{Length of pool (w. path)} & = (20 + 2x) \text{ m} \\ \\ \text{Breadth of pool (w. path)} & = (16 + 2x) \text{ m} \\ \\ \text{Area of pool (w. path)} & = \text{Area of pool} + \text{Area of path} \\ (20 + 2x)(16 + 2x) & = 20 \times 16 + 100 \\ 320 + 40x + 32x + 4x^2 & = 420 \\ 4x^2 + 72x + 320 & = 420 \\ 4x^2 + 72x + 320 - 420 & = 0 \\ 4x^2 + 72x - 100 & = 0 \\ x^2 + 18x - 25 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-18 \pm \sqrt{(18)^2 - 4(1)(-25)} \over 2(1)} \\ & = {-18 \pm \sqrt{424} \over 2} \\ & = 1.2956 \text{ or } -19.296 \text{ (Reject, since } x > 0) \\ \\ \text{Width of path} & = x \\ & = 1.2956 \\ & \approx 1.30 \text{ m} \end{align}
Try it yourself 19
(a)
\begin{align} \text{Time} & = {\text{Distance} \over \text{Speed}} \\ \\ \text{Time taken at original speed} & = {60 \over v} \text{ h} \\ \\ \text{Time taken w. faster speed} & = {60 \over v + 3} \text{ h} \\ \\ 40 \text{ mins} & = {40 \over 60} \text{ h} \\ & = {2 \over 3} \text{ h} \\ \\ {60 \over v} - {60 \over v + 3} & = {2 \over 3} \\ {60(3)(v + 3) \over 3v(v + 3)} - {60(3)(v) \over 3v(v + 3)} & = {2(v)(v + 3) \over 3v(v + 3)} \\ {180(v + 3) \over 3v(v + 3)} - {180 v \over 3v(v + 3)} & = {2v(v + 3) \over 3v(v + 3)} \\ {180(v + 3) - 180v \over 3v(v + 3)} & = {2v(v + 3) \over 3v(v + 3)} \\ \\ 180(v + 3) - 180v & = 2v(v + 3) \\ 180v + 540 - 180v & = 2v^2 + 6v \\ 540 & = 2v^2 + 6v \\ 0 & = 2v^2 + 6v - 540 \\ 0 & = v^2 + 3v - 270 \phantom{00} \text{ (Shown)} \end{align}
(b)
\begin{align} 0 & = v^2 + 3v - 270 \\ 0 & = (v - 15)(v + 18) \end{align} \begin{align} v - 15 & = 0 && \text{ or }& v + 18 & = 0 \\ v & = 15 &&& v & = -18 \text{ (Reject, since } v > 0) \end{align}
Try it yourself 20
(a)
| t | 0 | 0.5 | 1 | 1.5 | 2 | 2.5 | 3 |
|---|---|---|---|---|---|---|---|
| y | 15 | 14 | 12 | 9 | 5 | 0 | -6 |
(b)
$$ \text{No. From the graph, the maximum value of } y \text{ is 15, so } y \text{ cannot take values greater than 15} $$
(c)(i) The parcel starts at B (t = 0) and stops at A (t = 2.5)
$$ \text{Distance} = 15 \text{ m} $$
(c)(ii)
$$ \text{Time} = 2.5 \text{ s} $$
(d)
\begin{align} & T = 2.5, \text{ since the parcel reaches } A (y = 0) \text{ at } t = 2.5 \\ \\ & \text{The parcel is not expected to travel according to the same equation after it reaches } A \end{align}
Try it yourself 21
\begin{align} y & = (x + 2)(x - 1) \\ y & = x^2 - x + 2x - 2 \\ y & = x^2 + x - 2 \phantom{000000} [\text{Minimum curve } \cup] \\ \\ \text{Let } & x = 0, \\ y & = (0 + 2)(0 - 1) \\ y & = -2 \phantom{000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = (x + 2)(x - 1) \\ \\ x + 2 = 0 \phantom{0.} & \text{ or } \phantom{0} x - 1 = 0 \\ x = -2 & \phantom{0000000} x = 1 \phantom{000000} [x \text{-intercepts}] \\ \\ \text{Line of symmetry, } x & = {-2 + 1 \over 2} \\ x & = -0.5 \\ \\ \text{Let } & x = -0.5, \\ y & = (-0.5 + 2)(-0.5 - 1) \\ y & = -2.25 \\ \\ \text{Turning point: } & (-0.5, -2.25) \end{align}
Try it yourself 22
(a)
\begin{align} y & = 60x - x^2 \\ y & = - x^2 + 60x \phantom{000000} [\text{Maximum curve } \cap] \\ \\ \text{Let } & x = 0, \\ y & = - (0)^2 + 60(0) \\ y & = 0 \phantom{00000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = - x^2 + 60x \\ 0 & = x(-x + 60) \\ \\ x = 0 \phantom{0.} & \text{ or } \phantom{0} -x + 60 = 0 \\ & \phantom{00000000} - x = -60 \\ & \phantom{000000000-} x = 60 \phantom{000000} [x \text{-intercepts}] \\ \\ \text{Line of symmetry, } x & = {0 + 60 \over 2} \\ & = 30 \\ \\ \text{Let } & x = 30, \\ y & = -(30)^2 + 60(30) \\ y & = 900 \\ \\ \text{Turning point: } & (30, 900) \end{align}
(b)
\begin{align} \textbf{Maximum} \text{ point: } & (30, 900) \\ \\ \text{Vases sold } (x) & = 30 \\ \\ \text{Max. profit } (y) & = \$ 900 \end{align}
Try it yourself 23
\begin{align} y & = (x - 2)^2 + 3 \phantom{000000} [\text{Minimum curve } \cup] \\ \\ \text{Turning} & \text{ point: } (2, 3) \\ \\ \text{Let } & x = 0, \\ y & = (0 - 2)^2 + 3 \\ y & = 7 \phantom{000000000000000} [y \text{-intercept}] \end{align}
Try it yourself 24
(a)
\begin{align} -x^2 + 6x + 7 & = - (x^2 - 6x) + 7 \\ & = - \left[ x^2 - 6x + \left(6 \over 2\right)^2 - \left(6 \over 2\right)^2 \right] + 7 \phantom{000000} [\text{Complete the square}] \\ & = - ( x^2 - 6x + 3^2 - 3^2 ) + 7 \\ & = - [ (x - 3)^2 - 9] + 7 \\ & = - (x - 3)^2 + 9 + 7 \\ & = - (x - 3)^2 + 16 \end{align}
(b)
\begin{align} y & = -x^2 + 6x + 7 \\ y & = -(x - 3)^2 + 16 \phantom{000000} [\text{Maximum curve } \cap] \\ \\ \text{Turning} & \text{ point: } (3, 16) \\ \\ \text{Let } & x = 0, \\ y & = -(0 - 3)^2 + 16 \\ y & = 7 \phantom{00000000000000000.} [y \text{-intercept}] \end{align}
Ex 1.1: Solving Quadratic Equations by Completing the Square
Question 1
(a)
\begin{align} x^2 - 4x + 3 & = 0 \\ (x - 1)(x - 3) & = 0 \end{align} \begin{align} x - 1 & = 0 && \text{ or } & x - 3 & = 0 \\ x & = 1 &&& x & = 3 \end{align}
(b)
\begin{align} x^2 - 2x - 8 & = 0 \\ (x + 2)(x - 4) & = 0 \end{align} \begin{align} x + 2 & = 0 && \text{ or } & x - 4 & = 0 \\ x & = -2 &&& x & = 4 \end{align}
(c)
\begin{align} 2x^2 - 3x + 1 & = 0 \\ (2x - 1)(x - 1) & = 0 \end{align} \begin{align} 2x - 1 & = 0 && \text{ or } & x - 1 & = 0 \\ 2x & = 1 &&& x & = 1 \\ x & = {1 \over 2} \end{align}
(d)
\begin{align} 15x^2 + 8x + 1 & = 0 \\ (5x + 1)(3x + 1) & = 0 \end{align} \begin{align} 5x + 1 & = 0 && \text{ or } & 3x + 1 & = 0 \\ 5x & = -1 &&& 3x & = -1 \\ x & = - {1 \over 5} &&& x & = -{1 \over 3} \end{align}
(e)
\begin{align} 2x^2 + x - 6 & = 0 \\ (2x - 3)(x + 2) & = 0 \end{align} \begin{align} 2x - 3 & = 0 && \text{ or } & x + 2 & = 0 \\ 2x & = 3 &&& x & = -2 \\ x & = {3 \over 2} \end{align}
(f)
\begin{align} 3x^2 - 2x - 5 & = 0 \\ (3x - 5)(x + 1) & = 0 \end{align} \begin{align} 3x - 5 & =0 && \text{ or } & x + 1 & = 0 \\ 3x & = 5 &&& x & = -1 \\ x & = {5 \over 3} \end{align}
(g)
\begin{align} 4x^2 - 12x + 9 & = 0 \\ (2x - 3)(2x - 3) & = 0 \\ (2x - 3)^2 & = 0 \\ 2x - 3 & = 0 \\ 2x & = 3 \\ x & = {3 \over 2} \end{align}
(h)
\begin{align} 25x^2 - 16 & = 0 \\ (5x)^2 - (4)^2 & = 0 \\ (5x + 4)(5x - 4) & = 0 \phantom{000000} [a^2 - b^2 = (a + b)(a - b)] \end{align} \begin{align} 5x + 4 & = 0 && \text{ or } & 5x - 4 & = 0 \\ 5x & = -4 &&& 5x & = 4 \\ x & = -{4 \over 5} &&& x & = {4 \over 5} \end{align}
(i)
\begin{align} 14x^2 - 29x & = 15 \\ 14x^2 - 29x - 15 & = 0 \\ (2x - 5)(7x + 3) & = 0 \end{align} \begin{align} 2x - 5 & = 0 && \text{ or } & 7x + 3 & = 0 \\ 2x & = 5 &&& 7x & = -3 \\ x & = {5 \over 2} &&& x & = -{3 \over 7} \end{align}
(j)
\begin{align} x^2 - 2x & = 5x - 12 \\ x^2 - 2x - 5x + 12 & = 0 \\ x^2 - 7x + 12 & = 0 \\ (x - 3)(x - 4) & = 0 \end{align} \begin{align} x - 3 & = 0 && \text{ or } & x - 4 & = 0 \\ x & = 3 &&& x & = 4 \end{align}
Question 2
(a)
\begin{align} k & = \left(2 \over 2\right)^2 \\ & = 1^2 \\ & = 1 \end{align}
(b)
\begin{align} k & = \left(4 \over 2\right)^2 \\ & = 2^2 \\ & = 4 \end{align}
(c)
\begin{align} k & = \left(8 \over 2\right)^2 \\ & = 4^2 \\ & = 16 \end{align}
(d)
\begin{align} k & = \left(7 \over 2\right)^2 \\ & = {49 \over 4} \end{align}
(e)
\begin{align} k & = \left(1 \over 2\right)^2 \\ & = {1 \over 4} \end{align}
(f)
\begin{align} k & = \left(11 \over 2\right)^2 \\ & = {121 \over 4} \end{align}
Question 3
(a)
\begin{align} x^2 + 2x + 5 & = x^2 + 2x + \left(2 \over 2\right)^2 - \left(2 \over 2\right)^2 + 5 \\ & = x^2 + 2x + 1^2 - 1^2 + 5 \\ & = (x + 1)^2 - 1 + 5 \\ & = (x + 1)^2 + 4 \end{align}
(b)
\begin{align} x^2 - 6x - 3 & = x^2 - 6x + \left(6 \over 2\right)^2 - \left(6 \over 2\right)^2 - 3 \\ & = x^2 - 6x + 3^2 - 3^2 - 3 \\ & = (x - 3)^2 - 9 - 3 \\ & = (x - 3)^2 - 12 \end{align}
(c)
\begin{align} x^2 + 5x - 6 & = x^2 + 5x + \left(5 \over 2\right)^2 - \left(5 \over 2\right)^2 - 6 \\ & = \left(x + {5 \over 2}\right)^2 - {25 \over 4} - 6 \\ & = \left(x + {5 \over 2}\right)^2 - {49 \over 4} \end{align}
(d)
\begin{align} x^2 - 9x + 4 & = x^2 - 9x + \left(9 \over 2\right)^2 - \left(9 \over 2\right)^2 + 4 \\ & = \left(x - {9 \over 2}\right)^2 - {81 \over 4} + 4 \\ & = \left(x - {9 \over 2}\right)^2 - {65 \over 4} \end{align}
Question 4
(a)
\begin{align} (x - 1)^2 & = 25 \\ x - 1 & = \pm \sqrt{25} \\ x - 1 & = \pm 5 \end{align} \begin{align} x - 1 & = 5 && \text{ or } & x - 1 & = - 5 \\ x & = 5 + 1 &&& x & = -5 + 1 \\ x & = 6 &&& x & = -4 \end{align}
(b)
\begin{align} (x + 3)^2 & = 49 \\ x + 3 & = \pm \sqrt{49} \\ x + 3 & = \pm 7 \end{align} \begin{align} x + 3 & = 7 && \text{ or } & x + 3 & = -7 \\ x & = 7 - 3 &&& x & = -7 - 3 \\ x & = 4 &&& x & = -10 \end{align}
(c)
\begin{align} \left(x + {1 \over 2}\right)^2 & = 1 \\ x + {1 \over 2} & = \pm \sqrt{1} \\ x + {1 \over 2} & = \pm 1 \end{align} \begin{align} x + {1 \over 2} & = 1 && \text{ or } & x + {1 \over 2} & = -1 \\ x & = 1 - {1 \over 2} &&& x & = -1 - {1 \over 2} \\ x & = {1 \over 2} &&& x & = -{3 \over 2} \end{align}
(d)
\begin{align} \left(x - {5 \over 2}\right)^2 & = {9 \over 4} \\ x - {5 \over 2} & = \pm \sqrt{9 \over 4} \\ x - {5 \over 2} & = \pm {3 \over 2} \end{align} \begin{align} x - {5 \over 2} & = {3 \over 2} && \text{ or } & x - {5 \over 2} & = -{3 \over 2} \\ x & = {3 \over 2} + {5 \over 2} &&& x & = -{3 \over 2} + {5 \over 2} \\ x & = 4 &&& x & = 1 \end{align}
(e)
\begin{align} (x + 4)^2 & = 7 \\ x + 4 & = \pm \sqrt{7} \end{align} \begin{align} x + 4 & = \sqrt{7} && \text{ or } & x + 4 & = - \sqrt{7} \\ x & = \sqrt{7} - 4 &&& x & = -\sqrt{7} - 4 \\ x & \approx -1.35 &&& x & \approx -6.65 \end{align}
(f)
\begin{align} (x - 3)^2 & = 10 \\ x - 3 & = \pm \sqrt{10} \end{align} \begin{align} x - 3 & = \sqrt{10} && \text{ or } & x - 3 & = - \sqrt{10} \\ x & = \sqrt{10} + 3 &&& x & = -\sqrt{10} + 3 \\ x & \approx 6.16 &&& x & \approx -0.162 \end{align}
Question 5
(a)
\begin{align} x^2 + 8x + 12 & = x^2 + 8x + \left(8 \over 2\right)^2 - \left(8 \over 2\right)^2 + 12 \\ & = x^2 + 8x + 4^2 - 4^2 + 12 \\ & = (x + 4)^2 - 16 + 12 \\ & = (x + 4)^2 - 4 \end{align}
(b)
\begin{align} x^2 + 8x + 12 & = 0 \\ \underbrace{(x + 4)^2 - 4}_\text{From (a)} & = 0 \\ (x + 4)^2 & = 4 \\ x + 4 & = \pm \sqrt{4} \\ x + 4 & = \pm 2 \end{align} \begin{align} x + 4 & = 2 && \text{ or } & x + 4 & = -2 \\ x & = 2 - 4 &&& x & = -2 - 4 \\ x & = -2 &&& x & = -6 \end{align}
Question 6
(a)
\begin{align} x^2 - 6x + 4 & = x^2 - 6x + \left(6 \over 2\right)^2 - \left(6 \over 2\right)^2 + 4 \\ & = x^2 - 6x + 3^2 - 3^2 + 4 \\ & = (x - 3)^2 - 9 + 4 \\ & = (x - 3)^2 - 5 \end{align}
(b)
\begin{align} x^2 - 6x + 4 & = 0 \\ \underbrace{(x - 3)^2 - 5}_\text{From (a)} & = 0 \\ (x - 3)^2 & = 5 \\ x - 3 & = \pm \sqrt{5} \end{align} \begin{align} x - 3 & = \sqrt{5} && \text{ or } & x - 3 & = - \sqrt{5} \\ x & = \sqrt{5} + 3 &&& x & = - \sqrt{5} + 3 \\ x & \approx 5.24 &&& x & \approx 0.764 \end{align}
Question 7
(a)
\begin{align} x^2 + 2x + 3 & = x^2 + 2x + \left(2 \over 2\right)^2 - \left(2 \over 2\right)^2 + 3 \\ & = x^2 + 2x + 1^2 - 1^2 + 3 \\ & = (x + 1)^2 - 1 + 3 \\ & = (x + 1)^2 + 2 \end{align}
(b)
\begin{align} x^2 + 2x + 3 & = 0 \\ \underbrace{ (x + 1)^2 + 2 }_\text{From (a)} & = 0 \\ (x + 1)^2 & = -2 \\ x + 1 & = \pm \sqrt{-2} \\ \\ \text{No real roots } & \text{for } \sqrt{-2} \\ \\ \therefore \text{Equation has } & \text{no real roots} \end{align}
Question 8
(a)
\begin{align} x^2 + 4x + 2 & = 0 \\ x^2 + 4x + \left(4 \over 2\right)^2 - \left(4 \over 2\right)^2 + 2 & =0 \\ x^2 + 4x + 2^2 - 2^2 + 2 & = 0 \\ (x + 2)^2 - 4 + 2 & = 0 \\ (x + 2)^2 - 2 & = 0 \\ (x + 2)^2 & = 2 \\ x + 2 & = \pm \sqrt{2} \end{align} \begin{align} x + 2 & = \sqrt{2} && \text{ or } & x + 2 & = -\sqrt{2} \\ x & = \sqrt{2} - 2 &&& x & = - \sqrt{2} - 2 \\ x & \approx -0.586 &&& x & \approx -3.41 \end{align}
(b)
\begin{align} x^2 - 8x - 12 & = 0 \\ x^2 - 8x + \left(8 \over 2 \right)^2 - \left(8 \over 2\right)^2 - 12 & = 0 \\ x^2 - 8x + 4^2 - 4^2 - 12 & = 0 \\ (x - 4)^2 - 16 - 12 & = 0 \\ (x - 4)^2 - 28 & = 0 \\ (x - 4)^2 & = 28 \\ x - 4 & = \pm \sqrt{28} \end{align} \begin{align} x - 4 & = \sqrt{28} && \text{ or } & x - 4 & = - \sqrt{28} \\ x & = \sqrt{28} + 4 &&& x & = - \sqrt{28} + 4 \\ x & \approx 9.29 &&& x & \approx -1.29 \end{align}
(c)
\begin{align} x^2 - 7x + 13 & = 0 \\ x^2 - 7x + \left(7 \over 2\right)^2 - \left(7 \over 2\right)^2 + 13 & =0 \\ \left(x - {7 \over 2}\right)^2 - {49 \over 4} + 13 & = 0 \\ \left(x - {7 \over 2}\right)^2 + {3 \over 4} & = 0 \\ \left(x - {7 \over 2}\right)^2 & = - {3 \over 4} \\ x - {7 \over 2} & = \pm \sqrt{-{3 \over 4}} \\ \\ \text{No real roots } & \text{for } \sqrt{-{3 \over 4}} \\ \\ \therefore \text{Equation has } & \text{no real roots} \end{align}
(d)
\begin{align} x^2 + 5x + 7 & = 0 \\ x^2 + 5x + \left(5 \over 2\right)^2 - \left(5 \over 2\right)^2 + 7 & = 0 \\ \left(x + {5 \over 2}\right)^2 - {25 \over 4} + 7 & = 0 \\ \left(x + {5 \over 2}\right)^2 + {3 \over 4} & = 0 \\ \left(x + {5 \over 2}\right)^2 & = -{3 \over 4} \\ x + {5 \over 2} & = \pm \sqrt{-{3 \over 4}} \\ \\ \text{No real roots } & \text{for } \sqrt{-{3 \over 4}} \\ \\ \therefore \text{Equation has } & \text{no real roots} \end{align}
Question 9
(a)
\begin{align} x^2 - 12x + 26 & = 0 \\ x^2 - 12x + \left(12 \over 2\right)^2 - \left(12 \over 2\right)^2 + 26 & = 0 \\ x^2 - 12x + 6^2 - 6^2 + 26 & = 0 \\ (x - 6)^2 - 36 + 26 & = 0 \\ (x - 6)^2 - 10 & = 0 \\ (x - 6)^2 & = 10 \\ x - 6 & = \pm \sqrt{10} \end{align} \begin{align} x - 6 & = \sqrt{10} && \text{ or } & x - 6 & = - \sqrt{10} \\ x & = \sqrt{10} + 6 &&& x & = - \sqrt{10} + 6 \\ x & \approx 9.16 &&& x & \approx 2.84 \end{align}
(b)
\begin{align} x^2 - 9x - 21 & = 0 \\ x^2 - 9x + \left(9 \over 2\right)^2 - \left(9 \over 2\right)^2 - 21 & = 0 \\ \left(x - {9 \over 2}\right)^2 - {81 \over 4} - 21 & = 0 \\ \left(x - {9 \over 2}\right)^2 - {165 \over 4} & = 0 \\ \left(x - {9 \over 2}\right)^2 & = {165 \over 4} \\ x - {9 \over 2} & = \pm \sqrt{165 \over 4} \end{align} \begin{align} x - {9 \over 2} & = \sqrt{165 \over 4} && \text{ or } & x - {9 \over 2} & = -\sqrt{165 \over 4} \\ x & = \sqrt{165 \over 4} + {9 \over 2} &&& x & = -\sqrt{165 \over 4} + {9 \over 2} \\ x & \approx 10.92 &&& x & \approx -1.92 \end{align}
(c)
\begin{align} x^2 + 7x + 13 & = 0 \\ x^2 + 7x + \left(7 \over 2\right)^2 - \left(7 \over 2\right)^2 + 13 & = 0 \\ \left(x + {7 \over 2}\right)^2 - {49 \over 4} + 13 & = 0 \\ \left(x + {7 \over 2}\right)^2 + {3 \over 4} & = 0 \\ \left(x + {7 \over 2}\right)^2 & = -{3 \over 4} \\ x + {7 \over 2} & = \pm \sqrt{-{3 \over 4}} \\ \\ \text{No real roots } & \text{for } \sqrt{-{3 \over 4}} \\ \\ \therefore \text{Equation has } & \text{no real roots} \end{align}
(d)
\begin{align} x^2 - x - 1 & = 0 \\ x^2 - x + \left(1 \over 2\right)^2 - \left(1 \over 2\right)^2 - 1 & = 0 \\ \left(x - {1 \over 2}\right)^2 - {1 \over 4} - 1 & = 0 \\ \left(x - {1 \over 2}\right)^2 - {5 \over 4} & = 0 \\ \left(x - {1 \over 2}\right)^2 & = {5 \over 4} \\ x - {1 \over 2} & = \pm \sqrt{5 \over 4} \end{align} \begin{align} x - {1 \over 2} & = \sqrt{5 \over 4} && \text{ or } & x - {1 \over 2} & = -\sqrt{5 \over 4} \\ x & = \sqrt{5 \over 4} + {1 \over 2} &&& x & = -\sqrt{5 \over 4} + {1 \over 2} \\ x & \approx 1.62 &&& x & \approx -0.62 \end{align}
Question 10
(a)
\begin{align} x(x + 11) & = 2(x + 5) \\ x^2 + 11x & = 2x + 10 \\ x^2 + 11x - 2x - 10 & = 0 \\ x^2 + 9x - 10 & = 0 \\ (x + 10)(x - 1) & = 0 \end{align} \begin{align} x + 10 & = 0 && \text{ or } & x - 1 & = 0 \\ x & = -10 &&& x & = 1 \end{align}
(b)
\begin{align} (x + 2)(x - 3) & = 6 \\ x^2 - 3x + 2x - 6 & = 6 \\ x^2 - x - 6 & = 6 \\ x^2 - x - 6 - 6 & = 0 \\ x^2 - x - 12 & = 0 \\ (x + 3)(x - 4) & = 0 \end{align} \begin{align} x + 3 & = 0 && \text{ or } & x - 4 & = 0 \\ x & = -3 &&& x & = 4 \end{align}
(c)
\begin{align} (x + 1)(x - 7) & = 9 \\ x^2 - 7x + x - 7 & = 9 \\ x^2 - 6x - 7 & = 9 \\ x^2 - 6x - 7 - 9 & = 0 \\ x^2 - 6x - 16 & = 0 \\ (x + 2)(x - 8) & = 0 \end{align} \begin{align} x + 2 & = 0 && \text{ or } & x - 8 & = 0 \\ x & = -2 &&& x & = 8 \end{align}
(d)
\begin{align} 3x(2x + 5) & = 4(2x + 5) \\ 6x^2 + 15x & = 8x + 20 \\ 6x^2 + 15x - 8x - 20 & = 0 \\ 6x^2 + 7x - 20 & = 0 \\ (3x - 4)(2x + 5) & = 0 \end{align} \begin{align} 3x - 4 & = 0 && \text{ or } & 2x + 5 & = 0 \\ 3x & = 4 &&& 2x & = -5 \\ x & = {4 \over 3} &&& x & = -{5 \over 2} \end{align}
(e)
\begin{align} 36 - 25x^2 & = 0 \\ (6)^2 - (5x)^2 & = 0 \\ (6 + 5x)(6 - 5x) & = 0 \phantom{000000} [a^2 - b^2 = (a + b)(a - b)] \end{align} \begin{align} 6 + 5x & = 0 && \text{ or } & 6 - 5x & = 0 \\ 5x & = -6 &&& -5x & = -6 \\ x & = -{6 \over 5} &&& x & = {-6 \over -5} \\ & &&& x & = {6 \over 5} \end{align}
(f)
\begin{align} (7x + 2)(2x - 3) & = 5x - 6 \\ 14x^2 - 21x + 4x - 6 & = 5x - 6 \\ 14x^2 - 17x - 6 & = 5x - 6 \\ 14x^2 - 17x - 5x - 6 + 6 & = 0 \\ 14x^2 - 22x & = 0 \\ 2x (7x - 11) & = 0 \end{align} \begin{align} 2x & = 0 && \text{ or } & 7x - 11 & = 0 \\ x & = 0 &&& 7x & = 11 \\ & &&& x & = {11 \over 7} \end{align}
(g)
\begin{align} (x + 1)^2 & = 9(x + 1) \\ (x + 1)(x + 1) & = 9(x + 1) \\ x^2 + x + x + 1 & = 9x + 9 \\ x^2 + 2x + 1 & = 9x + 9 \\ x^2 + 2x - 9x + 1 - 9 & = 0 \\ x^2 - 7x - 8 & = 0 \\ (x + 1)(x - 8) & = 0 \end{align} \begin{align} x + 1 & = 0 && \text{ or } & x - 8 & = 0 \\ x & = -1 &&& x & = 8 \end{align}
(h)
\begin{align} 2(3x + 4)^2 - 5(3x + 4) + 3 & = 0 \\ \\ \text{Let } y = 3x + 4, & \\ 2y^2 - 5y + 3 & = 0 \\ (2y - 3)(y - 1) & = 0 \end{align} \begin{align} 2y - 3 & = 0 && \text{ or } & y - 1 & = 0 \\ 2y & = 3 &&& y & = 1 \\ y & = {3 \over 2} \\ \\ \therefore 3x + 4 & = {3 \over 2} &&& 3x + 4 & = 1 \\ 3x & = {3 \over 2} - 4 &&& 3x & = 1 - 4 \\ 3x & = -{5 \over 2} &&& 3x & = -3 \\ x & = -{5 \over 2} \div 3 &&& x & = {-3 \over 3} \\ x & = -{5 \over 6} &&& x & = -1 \end{align}
Question 11
(a)
\begin{align} 3x^2 + kx - 5 & = 0 \\ \\ \text{Let } & x = -5, \\ 3(-5)^2 + k(-5) - 5 & = 0 \\ 3(25) - 5k - 5 & = 0 \\ 75 - 5k - 5 & = 0 \\ 70 - 5k & = 0 \\ -5k & = -70 \\ k & = {-70 \over -5} \\ k & = 14 \end{align}
(b)
\begin{align}
3x^2 + kx - 5 & = 0 \\
3x^2 + (14)x - 5 & = 0 \\
3x^2 + 14x - 5 & = 0 \\
(x + 5)(3x - 1) & = 0
\end{align}
\begin{align}
x + 5 & = 0
&& \text{ or } &
3x - 1 & = 0 \\
x & = -5
&&&
3x & = 1 \\
& &&&
x & = {1 \over 3}
\end{align}
$$ \text{Other root is } x = {1 \over 3} $$
Question 12
(a)
\begin{align} 2x^2 + kx - k^2 & = 0 \\ \\ \text{Let } & x = -3, \\ 2(-3)^2 + k(-3) - k^2 & = 0 \\ 2(9) - 3k - k^2 & = 0 \\ 18 - 3k - k^2 & = 0 \\ (6 + k)(3 - k) & = 0 \end{align} \begin{align} 6 + k & = 0 && \text{ or } & 3 - k & = 0 \\ k & = -6 &&& -k & = -3 \\ & &&& k & = 3 \end{align}
(b)
\begin{align}
\text{For } k = -6, & \\
2x^2 + (-6)x - (-6)^2 & = 0 \\
2x^2 - 6x - 36 & = 0 \\
x^2 - 3x - 18 & = 0\phantom{000000} [\div 2] \\
(x + 3)(x - 6) & = 0
\end{align}
\begin{align}
x + 3 & = 0
&& \text{ or } &
x - 6 & = 0 \\
x & = - 3
&&&
x & = 6
\end{align}
$$ \text{For } k = - 6, \text{ other root is } x = 6 $$
\begin{align}
\text{For } k = 3, & \\
2x^2 + (3)x - (3)^2 & = 0 \\
2x^2 + 3x - 9 & = 0 \\
(x + 3)(2x - 3) & = 0
\end{align}
\begin{align}
x + 3 & = 0
&& \text{ or } &
2x - 3 & = 0 \\
x & = - 3
&&&
2x & = 3 \\
& &&&
x & = {3 \over 2}
\end{align}
$$ \text{For } k = 3, \text{ other root is } x = {3 \over 2} $$
Question 13
(a)
\begin{align} -x^2 + 6x + 5 & = 0 \\ x^2 - 6x - 5 & = 0 \\ x^2 - 6x + \left(6 \over 2\right)^2 - \left(6 \over 2\right)^2 - 5 & = 0 \phantom{000000} [\text{Complete the square}] \\ x^2 - 6x + 3^2 - 3^2 - 5 & = 0 \\ (x - 3)^2 - 9 - 5 & = 0 \\ (x - 3)^2 - 14 & = 0 \\ (x - 3)^2 & = 14 \\ x - 3 & = \pm \sqrt{14} \end{align} \begin{align} x - 3 & = \sqrt{14} && \text{ or } & x - 3 & = -\sqrt{14} \\ x & = \sqrt{14} + 3 &&& x & = -\sqrt{14} + 3 \\ x & \approx 6.74 &&& x & \approx -0.742 \end{align}
(b)
\begin{align} 2x^2 - 8x + 14 & = 0 \\ x^2 - 4x + 7 & = 0 \\ x^2 - 4x + \left(4 \over 2\right)^2 - \left(4 \over 2 \right)^2 + 7 & = 0 \phantom{000000} [\text{Complete the square}] \\ x^2 - 4x + 2^2 - 2^2 + 7 & = 0 \\ (x - 2)^2 - 4 + 7 & = 0 \\ (x - 2)^2 + 3 & = 0 \\ (x - 2)^2 & = -3 \\ x - 2 & = \pm \sqrt{-3} \\ \\ \text{No real roots } & \text{for } \sqrt{-3} \\ \\ \therefore \text{Equation has } & \text{no real roots} \end{align}
Question 14
(a)
\begin{align} x & = - 3 && \text{ or } & x & = 5 \\ x + 3 & = 0 &&& x - 5 & = 0 \end{align} \begin{align} (x + 3)(x - 5) & = 0 \\ x^2 - 5x + 3x - 15 & = 0 \\ x^2 - 2x - 15 & =0 \\ \\ \therefore b = -2, c & = -15 \end{align}
(b)
\begin{align} x & = -p && \text{ or } & x & = q \\ x + p & = 0 &&& x - q & = 0 \end{align} \begin{align} (x + p)(x - q) & = 0 \\ x^2 - qx + px - pq & = 0 \\ x^2 + px - qx - pq & = 0 \\ x^2 + (p - q)x - pq & = 0 \\ \\ \therefore b = p - q, c & = -pq \end{align}
Question 15
(a)
\begin{align} 9x^2 - 12xy + 4y^2 & = (3x - 2y)(3x - 2y) \\ & = (3x - 2y)^2 \end{align}
(b)
\begin{align} 9x^2 - 12xy + 4y^2 & = 0 \\ (3x - 2y)^2 & = 0 \\ 3x - 2y & = 0 \\ 3x & = 2y \\ {3x \over y} & = 2 \\ {x \over y} & = {2 \over 3} \\ 5 \left(x \over y\right) & = 5 \left(2 \over 3\right) \\ {5x \over y} & = {10 \over 3} \end{align}
Question 16
\begin{align}
3x(x + 4) & = x(x + 1) \\
3x^2 + 12x & = x^2 + x \\
3x^2 - x^2 + 12x - x & = 0 \\
2x^2 + 11x & = 0 \\
x(2x + 11) & = 0
\end{align}
\begin{align}
x & = 0
&& \text{ or } &
2x + 11 & = 0 \\
& &&&
2x & = -11 \\
& &&&
x & = -{11 \over 2}
\end{align}
$$ \text{No. Dividing both sides by } x \text{ eliminates the root } x = 0 $$
Question 17
\begin{align}
(5x - 2)^2 & = (2x + 1)^2 \\
5x - 2 & = \pm \sqrt{(2x + 1)^2} \\
5x - 2 & = \pm (2x + 1)
\end{align}
\begin{align}
5x - 2 & = 2x + 1
&& \text{ or } &
5x - 2 & = -(2x + 1) \\
5x - 2x & = 1 + 2
&&&
5x - 2 & = -2x - 1 \\
3x & = 3
&&&
5x + 2x & = -1 + 2 \\
x & = 1
&&&
7x & = 1 \\
& &&&
x & = {1 \over 7}
\end{align}
$$ \text{His solution is incomplete as he left out the equation } 5x - 2 = - (2x + 1) $$
Question 18
\begin{align} x & = 2 - \sqrt{3} && \text{ or } & x & = 2 + \sqrt{3} \\ x + \sqrt{3} - 2 & = 0 &&& x - \sqrt{3} - 2 & = 0 \end{align} \begin{align} (x + \sqrt{3} - 2)(x - \sqrt{3} - 2) & = 0 \\ x^2 - \sqrt{3} x - 2x + \sqrt{3} x - (\sqrt{3})(\sqrt{3}) - 2 \sqrt{3} - 2x + 2 \sqrt{3} + 4 & = 0 \\ x^2 - \sqrt{3} x - 2x + \sqrt{3} x - 3 - 2 \sqrt{3} - 2x + 2 \sqrt{3} + 4 & = 0 \\ x^2 - \sqrt{3} x + \sqrt{3}x - 2x - 2x - 3 - 2\sqrt{3} + 2 \sqrt{3} + 4 & = 0 \\ x^2 - 4x + 1 & = 0 \end{align}
Ex 1.2: Solving Quadratic Equations by Using the Quadratic Formula
Question 1
(a)
\begin{align} x^2 & - 12x + 36 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-12) \pm \sqrt{(-12)^2 - 4(1)(36)} \over 2(1)} \\ & = {12 \pm \sqrt{0} \over 2} \\ & = 6 \end{align}
(b)
\begin{align} 2x^2 & + 6x + 1 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-6 \pm \sqrt{(6)^2 - 4(2)(1)} \over 2(2)} \\ & = {-6 \pm \sqrt{28} \over 4} \\ & = -0.17712 \text{ or } -2.8228 \\ & \approx -0.18 \text{ or } -2.82 \text{ (2 d.p.)} \end{align}
(c)
\begin{align} 3x^2 & - 2x - 5 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {- (-2) \pm \sqrt{(-2)^2 - 4(3)(-5)} \over 2(3)} \\ & = {2 \pm \sqrt{64} \over 6} \\ & = {5 \over 3} \text{ or } -1 \\ & \approx 1.67 \text{ or } -1 \text{ (2 d.p.)} \end{align}
(d)
\begin{align} 5x^2 & - x + 4 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {- (-1) \pm \sqrt{ (-1)^2 - 4(5)(4)} \over 2(5)} \\ & = {1 \pm \sqrt{-79} \over 10} \\ \\ \\ \sqrt{-79} & \text{ is not a real number} \\ \\ \therefore \text{Equa} & \text{tion has no real roots} \end{align}
(e)
\begin{align} 7x^2 & + x - 2 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {- 1 \pm \sqrt{ (1)^2 - 4(7)(-2)} \over 2(7)} \\ & = {-1 \pm \sqrt{57} \over 14} \\ & = 0.46784 \text{ or } -0.6107 \\ & \approx 0.47 \text{ or } - 0.61 \text{ (2 d.p.)} \end{align}
Question 2
(a)
\begin{align} x + 7x(1 - x) & = 3x - 26 \\ x + 7x - 7x^2 & = 3x - 26 \\ 8x - 7x^2 & = 3x - 26 \\ 0 & = 7x^2 + 3x - 8x - 26 \\ 0 & = 7x^2 - 5x - 26 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {- (-5) \pm \sqrt{ (-5)^2 - 4(7)(-26)} \over 2(7)} \\ & = {5 \pm \sqrt{753} \over 14} \\ & = 2.3172 \text{ or } -1.6029 \\ & \approx 2.32 \text{ or } -1.60 \text{ (3 s.f.)} \end{align}
(b)
\begin{align} (4x - 5)(x + 2) & = 4x + 1 \\ 4x^2 + 8x - 5x - 10 & = 4x + 1 \\ 4x^2 + 3x - 10 & = 4x + 1 \\ 4x^2 + 3x - 4x - 10 - 1 & = 0 \\ 4x^2 - x - 11 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {- (-1) \pm \sqrt{(-1)^2 - 4(4)(-11)} \over 2(4)} \\ & = {1 \pm \sqrt{177} \over 8} \\ & = 1.788 \text{ or } -1.538 \\ & \approx 1.79 \text{ or } -1.54 \text{ (3 s.f.)} \end{align}
(c)
\begin{align} x(1 - 2x) + 9 & = 0 \\ x - 2x^2 + 9 & = 0 \\ - 2x^2 + x + 9 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {- 1 \pm \sqrt{(1)^2 - 4(-2)(9)} \over 2(-2)} \\ & = {-1 \pm \sqrt{73} \over -4} \\ & = -1.886 \text{ or } 2.386 \\ & \approx -1.89 \text{ or } 2.39 \text{ (3 s.f.)} \end{align}
(d)
\begin{align} (1 - x)(4 + x) & = 5 \\ 4 + x - 4x - x^2 & = 5 \\ 4 - 3x - x^2 & = 5 \\ 0 & = x^2 + 3x + 5 - 4 \\ 0 & = x^2 + 3x + 1 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-3 \pm \sqrt{(3)^2 - 4(1)(1)} \over 2(1)} \\ & = {-3 \pm \sqrt{5} \over 2} \\ & = -0.38196 \text{ or } -2.618 \\ & \approx -0.382 \text{ or } -2.62 \text{ (3 s.f.)} \end{align}
(e)
\begin{align} x^2 & = {3 \over 4}x + 2 \\ x^2 - {3 \over 4}x - 2 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {- \left(-{3 \over 4}\right) \pm \sqrt{ \left(-{3 \over 4}\right)^2 - 4(1)(-2)} \over 2(1)} \\ & = { {3 \over 4} \pm \sqrt{ 137 \over 16} \over 2 } \\ & = 1.838 \text{ or } -1.088 \\ & \approx 1.84 \text{ or } -1.09 \text{ (3 s.f.)} \end{align}
Question 3
(a)
\begin{align} (x - 2)(x + 1) & = 8 \\ x^2 + x - 2x - 2 & = 8 \\ x^2 - x - 2 & = 8 \\ x^2 - x - 2 - 8 & = 0 \\ x^2 - x - 10 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {- (-1) \pm \sqrt{(-1)^2 - 4(1)(-10)} \over 2(1)} \\ & = {1 \pm \sqrt{41} \over 2} \\ & = 3.7015 \text{ or } -2.7015 \\ & \approx 3.70 \text{ or } -2.70 \text{ (2 d.p.)} \end{align}
(b)
\begin{align} 3x^2 - 4 & = 7x \\ 3x^2 - 7x - 4 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-7) \pm \sqrt{ (-7)^2 - 4(3)(-4)} \over 2(3)} \\ & = {7 \pm \sqrt{97} \over 6} \\ & = 2.8081 \text{ or } -0.4748 \\ & \approx 2.81 \text{ or } -0.47 \text{ (2 d.p.)} \end{align}
(c)
\begin{align} (x - 2)(x + 2) & = 4(x - 2) \\ x^2 - 2^2 & = 4x - 8 \phantom{000000} [ (a - b)(a + b) = a^2 - b^2] \\ x^2 - 4 & = 4x - 8 \\ x^2 - 4x - 4 + 8 & = 0 \\ x^2 - 4x + 4 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-4) \pm \sqrt{ (-4)^2 - 4(1)(4)} \over 2(1)} \\ & = {4 \pm \sqrt{0} \over 2} \\ & = 2 \end{align}
(d)
\begin{align} 5x(x + 1) & = 3(x - 2) \\ 5x^2 + 5x & = 3x - 6 \\ 5x^2 + 5x - 3x + 6 & = 0 \\ 5x^2 + 2x + 6 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-2 \pm \sqrt{(2)^2 - 4(5)(6)} \over 2(5)} \\ & = {-2 \pm \sqrt{-116} \over 10} \\ \\ \\ \sqrt{-116} \text{ is } & \text{not a real number} \\ \\ \therefore \text{Equation } & \text{has no real roots} \end{align}
(e)
\begin{align} (2x + 1)^2 & = (x - 3)^2 \\ \underbrace{(2x)^2 + 2(2x)(1) + (1)^2}_{(a + b)^2 = a^2 + 2ab + b^2} & = \underbrace{(x)^2 - 2(x)(3) + (3)^2}_{(a - b)^2 = a^2 - 2ab + b^2} \\ 4x^2 + 4x + 1 & = x^2 - 6x + 9 \\ 4x^2 - x^2 + 4x + 6x + 1 - 9 & = 0 \\ 3x^2 + 10x - 8 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-10 \pm \sqrt{(10)^2 - 4(3)(-8)} \over 2(3)} \\ & = {-10 \pm \sqrt{196} \over 6} \\ & = {2 \over 3} \text{ or } -4 \end{align}
(f)
\begin{align} x(x - 6) & = {2 \over 3}(x^2 + 1) \\ x^2 - 6x & = {2 \over 3}x^2 + {2 \over 3} \\ x^2 - {2 \over 3}x^2 - 6x - {2 \over 3} & = 0 \\ {1 \over 3}x^2 - 6x - {2 \over 3} & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-6) \pm \sqrt{ (-6)^2 - 4 \left(1 \over 3\right) \left(-{2 \over 3}\right)} \over 2 \left(1 \over 3\right)} \\ & = {6 \pm \sqrt{ 332 \over 9} \over {2 \over 3}} \\ & = 18.110 \text{ or } -0.1104 \\ & \approx 18.11 \text{ or } -0.11 \text{ (2 d.p.)} \end{align}
Question 4
(a)
\begin{align}
{1 \over 2}x^2 & + 3x + q = 0 \\
\\
x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\
& = {-3 \pm \sqrt{(3)^2 - 4\left(1 \over 2\right)(q)} \over 2 \left(1 \over 2\right) } \\
& = {-3 \pm \sqrt{9 - 2q} \over 1} \\
& = {-3 \pm \sqrt{9 - 2q} } \\
\\ \\
\implies p + \sqrt{7} & = -3 + \sqrt{9 - 2q} \\
\\
\text{Comparing } & \text{both sides,}
\end{align}
\begin{align}
p & = -3
&&&
\sqrt{7} & = \sqrt{9 - 2q} \\
& &&&
7 & = 9 - 2q \\
& &&&
2q & = 9 - 7 \\
& &&&
2q & = 2 \\
& &&&
q & = 1
\end{align}
(b)
\begin{align} x & = -3 - \sqrt{9 - 2q} \\ & = -3 - \sqrt{9 - 2(1)} \\ & = -3 - \sqrt{7} \\ & \approx -5.65 \text{ (3 s.f.)} \end{align}
Question 5
Method 1: Quadratic formula
\begin{align} x^2 & - 6x + 14 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-6) \pm \sqrt{ (-6)^2 - 4(1)(14)} \over 2(1)} \\ & = {6 \pm \sqrt{-20} \over 2} \\ \\ \\ \sqrt{-20} &\text{ is } \text{not a real number} \\ \\ \therefore \text{Equa} & \text{tion has no real roots} \end{align}
Method 2: Complete the square
\begin{align} x^2 - 6x + 14 & = 0 \\ x^2 - 6x + \left(6 \over 2\right)^2 - \left(6 \over 2\right)^2 + 14 & = 0 \\ x^2 - 6x + 3^2 - 3^2 + 14 & = 0 \\ (x - 3)^2 - 9 + 14 & = 0 \\ (x - 3)^2 + 5 & = 0 \\ (x - 3)^2 & = -5 \\ \\ (x - 3)^2 \ge 0 \text{ for } & \text{all real values of } x \\ \\ \therefore \text{Equation } & \text{has no real roots} \end{align}
Ex 1.3: Solving Quadratic Equations by Using the Graphical Method
Question 1
(a)
| x | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|---|
| y | 9 | 4 | 1 | 0 | 1 | 4 | 9 |
(b)
(c) Look for the x-intercept of the graph
$$ \text{From graph, } x = 2 $$
Question 2
(a)
| x | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|---|
| y | -4 | 0 | 2 | 2 | 0 | -4 | -10 |
(b)
(c) Look for the x-intercepts of the graph
$$ \text{From graph, } x = 0, 3 $$
Question 3
(a)
| x | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|
| y | 10 | 5 | 2 | 1 | 2 | 5 | 10 |
(b)
(c) The graph does not meet the x-axis at all!
$$ \text{From graph, there are no real solutions} $$
Question 4
(a)
\begin{align} x^2 & = x + 6 \\ \underbrace{x^2 - x - 6}_\text{Curve} & = 0 \\ \\ \text{From graph, } & x = -2, 3 \phantom{000000} [\text{Look for } x \text{-intercepts}] \end{align}
(b)(i)
\begin{align} x^2 - x - 5 & = 0 \\ x^2 - x - 5 - 1 & = -1 \\ \underbrace{x^2 - x - 6}_\text{Curve} & = -1 \\ \\ \text{Draw } & y = -1 \end{align}
$$ \text{From graph, } x = -1.8, 2.8 $$
(b)(ii)
\begin{align} x^2 - 2x + 1 & = 0 \\ x^2 - 2x + x + 1 & = 0 + x \\ x^2 - x + 1 & = x \\ x^2 - x + 1 - 7 & = x - 7 \\ \underbrace{x^2 - x - 6}_\text{Curve} & = x - 7 \\ \\ \text{Draw } & y = x - 7 \end{align}
| x | 0 | 2 | 4 |
|---|---|---|---|
| y | -7 | -5 | -3 |
$$ \text{From graph, } x = 1 $$
Question 5
(a)
| x | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|
| y | 5 | 0 | -3 | -4 | -3 | 0 | 5 |
(b)
\begin{align} x^2 & = 2x + 3 \\ \underbrace{x^2 - 2x - 3}_\text{Curve} & = 0 \\ \\ \text{From graph, } & x = -1, 3 \phantom{000000} [\text{Look for } x \text{-intercepts}] \end{align}
(c)(i)
\begin{align} x^2 - 2x & = 6 \\ x^2 - 2x - 3 & = 6 - 3 \\ \underbrace{x^2 - 2x - 3}_\text{Curve} & = 3 \\ \\ \text{Draw } & y = 3 \\ \\ \text{From graph, } & x = -1.65, 3.65 \end{align}
(c)(ii)
\begin{align} x^2 - x - 3 & = 0 \\ x^2 - x - 3 - x & = -x \\ \underbrace{x^2 - 2x - 3}_\text{Curve} & = -x \\ \\ \text{Draw } & y = -x \end{align}
| x | 0 | 1 | 2 |
|---|---|---|---|
| y | 0 | -1 | -2 |
$$ \text{From graph, } x = -1.3, 2.3 $$
Question 6
(a)
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 |
|---|---|---|---|---|---|---|---|
| y | 11 | 2 | -3 | -4 | -1 | 6 | 17 |
(b)
\begin{align} 2x^2 + 5x & = 1 \\ \underbrace{2x^2 + 5x - 1}_\text{Curve} & = 0 \\ \\ \text{From graph, } & x = -2.7, 0.2 \phantom{000000} [\text{Look for } x \text{-intercepts}] \end{align}
(c)(i)
\begin{align} 2x^2 + 5x - 5 & = 0 \\ 2x^2 + 5x - 5 + 4 & = 0 + 4 \\ \underbrace{2x^2 + 5x - 1}_\text{Curve} & = 4 \\ \\ \text{Draw } & y = 4 \\ \\ \text{From graph, } & x = -3.25, 0.75 \end{align}
(c)(ii)
\begin{align} 2x^2 + x - 6 & = 0 \\ 2x^2 + x - 6 + 4x & = 0 + 4x \\ 2x^2 + 5x - 6 & = 4x \\ 2x^2 + 5x - 6 + 5 & = 4x + 5 \\ \underbrace{2x^2 + 5x - 1}_\text{Curve} & = 4x + 5 \\ \\ \text{Draw } & y = 4x + 5 \end{align}
| x | -1 | 0 | 1 |
|---|---|---|---|
| y | 1 | 5 | 9 |
$$ \text{From graph, } x = -2, 1.5 $$
Question 7
(a)
| x | -2 | -1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|
| y | 20 | 9 | 4 | 5 | 12 | 25 |
(b)
\begin{align} 3x^2 & = 2x - 4 \\ \underbrace{3x^2 - 2x + 4}_\text{Curve} & = 0 \\ \\ \text{From graph, no real solutions} & \text{ since there are no } x \text{-intercept(s)} \end{align}
(c)(i)
\begin{align} 3x^2 - 2x - 10 & = 0 \\ 3x^2 - 2x - 10 + 14 & = 0 + 14 \\ \underbrace{3x^2 - 2x + 4}_\text{Curve} & = 14 \\ \\ \text{Draw } & y = 14 \\ \\ \text{From graph, } & x = -1.5, 2.2 \end{align}
(c)(ii)
\begin{align} 3x^2 - 5 & = 0 \\ 3x^2 - 5 - 2x & = 0 - 2x \\ 3x^2 - 2x - 5 & = -2x \\ 3x^2 - 2x - 5 + 9 & = -2x + 9 \\ \underbrace{3x^2 - 2x + 4}_\text{Curve} & = -2x + 9 \\ \\ \text{Draw } & y = - 2x + 9 \end{align}
| x | 0 | 1 | 2 |
|---|---|---|---|
| y | 9 | 7 | 5 |
$$ \text{From graph, } x = -1.3, 1.3 $$
Question 8
(a)
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 |
|---|---|---|---|---|---|---|---|
| y | -2 | 3 | 6 | 7 | 6 | 3 | -2 |
(b)
\begin{align} x^2 + 2x - 6 & = 0 \\ 0 & = \underbrace{6 - 2x - x^2}_\text{Curve} \\ \\ \text{From graph, } & x = -3.65, 1.65 \phantom{000000} [\text{Look for } x \text{-intercepts}] \end{align}
(c)(i)
\begin{align} x^2 + 2x - 2 & = 0 \\ 0 & = 2 - 2x - x^2 \\ 0 + 4 & = 2 - 2x - x^2 + 4 \\ 4 & = \underbrace{6 - 2x - x^2}_\text{Curve} \\ \\ \text{Draw } & y = 4 \\ \\ \text{From graph, } & x = -2.75, 0.75 \end{align}
(c)(ii)
\begin{align} 3 - x - x^2 & = 0 \\ 3 - x - x^2 + 3 & = 0 + 3 \\ 6 - x - x^2 & = 3 \\ 6 - x - x^2 - x & = 3 - x \\ \underbrace{6 - 2x - x^2}_\text{Curve} & = 3 - x \\ \\ \text{Draw } & y = 3 - x \end{align}
| x | 0 | 1 | 2 |
|---|---|---|---|
| y | 3 | 2 | 1 |
$$ \text{From graph, } x = -2.3, 1.3 $$
Question 9
(a)
| x | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|
| y | -22 | -7 | 2 | 5 | 2 | -7 | -22 |
(b)(i)
\begin{align} 3x^2 - 6x - 2 & = 0 \\ 0 & = \underbrace{-3x^2 + 6x + 2}_\text{Curve} \\ \\ \text{From graph, } & x = -0.3, 2.3 \phantom{000000} [\text{Look for } x \text{-intercepts}] \end{align}
(b)(ii)
\begin{align} 7 + 6x - 3x^2 & = 0 \\ - 3x^2 + 6x + 7 & = 0 \\ - 3x^2 + 6x + 7 - 5 & = 0 - 5 \\ \underbrace{ -3x^2 + 6x + 2 }_\text{Curve} & = - 5 \\ \\ \text{Draw } & y = -5 \\ \\ \text{From graph, } & x = -0.8, 2.8 \end{align}
(c)(i)
\begin{align} 5x - y & = 4 \\ -y & = 4 - 5x \\ y & = -4 + 5x \end{align}
| x | 0 | 1 | 2 |
|---|---|---|---|
| y | -4 | 1 | 6 |
$$ \text{From graph, } x = -1.25, 1.6$$
(c)(ii)
\begin{align} y & = -4 + 5x \phantom{00} \text{--- (1)} \\ \\ y & = - 3x^2 + 6x + 2 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ -4 + 5x & = -3x^2 + 6x + 2 \\ 0 & = -3x^2 + 6x - 5x + 2 + 4 \\ 0 & = -3x^2 + x + 6 \\ \\ \therefore b & = 1, c = 6 \end{align}
Question 10
(a)
| x | -6 | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 |
|---|---|---|---|---|---|---|---|---|---|
| y | 6 | 2.5 | 0 | -1.5 | -2 | -1.5 | 0 | 2.5 | 6 |
(b)
\begin{align} {1 \over 2}x^2 + 2x + 1 & = 0 \\ \underbrace{ {1 \over 2}x^2 + 2x }_\text{Curve} & = -1 \\ \\ \text{Draw } & y = - 1 \\ \\ \text{From graph, } & x = -3.4, -0.6 \end{align}
(c)(i)
\begin{align} x + 2y & = 4 \\ 2y & = 4 - x \\ y & = {1 \over 2}(4 - x) \\ y & = 2 - {1 \over 2}x \end{align}
| x | 0 | 1 | 2 |
|---|---|---|---|
| y | 2 | 1.5 | 1 |
$$ \text{From graph, } x = -5.7, 0.7 $$
(c)(ii)
\begin{align} y & = {1 \over 2}x^2 + 2x \phantom{00} \text{--- (1)} \\ \\ y & = 2 - {1 \over 2}x \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ {1 \over 2}x^2 + 2x & = 2 - {1 \over 2}x \\ {1 \over 2}x^2 + 2x + {1 \over 2}x - 2 & = 0 \\ {1 \over 2}x^2 + {5 \over 2}x - 2 & = 0 \\ 2 \left({1 \over 2}x^2 + {5 \over 2}x - 2\right) & = 2(0) \\ x^2 + 5x - 4 & = 0 \end{align}
Question 11
(a)
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|
| y | -16 | -9 | -4 | -1 | 0 | -1 | -4 | -9 |
(b)
\begin{align} -2x^2 + 4x & = 2 \\ {1 \over 2}(-2x^2 + 4x) & = {1 \over 2}(2) \\ -x^2 + 2x & = 1 \\ \underbrace{-x^2 + 2x - 1}_\text{Curve} & = 0 \\ \\ \text{From graph, } & x = 1 \phantom{000000} [\text{Look for } x \text{-intercept}] \end{align}
(c) The maximum point of the graph is (1, 0). Thus, k must be smaller than 0,
$$ k < 0 $$
(d)
\begin{align} x^2 + 2x + 2 & = mx \\ -(x^2 + 2x + 2) & = - mx \\ - x^2 - 2x - 2 & = - mx \\ - x^2 - 2x - 2 + 4x & = -mx + 4x \\ - x^2 + 2x - 2 & = (4 - m)x \\ - x^2 + 2x - 2 + 1 & = (4 - m)x + 1 \\ \underbrace{-x^2 + 2x - 1}_\text{Curve} & = (4 - m)x + 1 \\ \\ \text{Draw } & y = (4 - m)x + 1 \\ \\ \text{Let } & x = 0, \\ y & = (4 - m)(0) + 1 \\ y & = 0 + 1 \\ y & = 1 \\ \\ \therefore \text{Construct two lines } & \text{from } (0, 1) \text{ that meet curve only once} \\ \\ \\ y & = (4 - m)x + 1 \end{align} \begin{align} \text{Using } & (-1.45, -6), &&& \text{Using } & (1.45, -0.2), \\ -6 & = (4 - m)(-1.45) + 1 &&& -0.2 & = (4 - m)(1.45) + 1 \\ -6 & = -5.8 + 1.45m + 1 &&& -0.2 & = 5.8 - 1.45m + 1 \\ -1.45m & = -5.8 + 1 + 6 &&& 1.45m & = 5.8 + 1 + 0.2 \\ -1.45m & = 1.2 &&& 1.45m & = 7 \\ m & = {1.2 \over -1.45} &&& m & = {7 \over 1.45} \\ m & = -0.82758 &&& m & = 4.8276 \\ m & \approx -0.83 \text{ (2 d.p.)} &&& m & \approx 4.83 \text{ (2 d.p.)} \end{align}
Ex 1.4: Solving Fractional Equations That Can Be Reduced to Quadratic Equations
Question 1
(a)
\begin{align} {x - 1 \over 2} & = {x \over x + 3} \\ {(x - 1)(x + 3) \over 2(x + 3)} & = {2x \over 2(x + 3)} \\ \\ (x - 1)(x + 3) & = 2x \\ x^2 + 3x - x - 3 & = 2x \\ x^2 + 2x - 3 & = 2x \\ x^2 + 2x - 2x - 3 & = 0 \\ x^2 - 3 & = 0 \\ x^2 & = 3 \\ x & = \pm \sqrt{3} \end{align}
(b)
\begin{align} {x - 1 \over 2x - 5} & = {x - 6 \over x + 2} \\ {(x - 1)(x + 2) \over (2x - 5)(x + 2)} & = {(x - 6)(2x - 5) \over (2x - 5)(x + 2)} \\ \\ (x - 1)(x + 2) & = (x - 6)(2x - 5) \\ x^2 + 2x - x - 2 & = 2x^2 - 5x - 12x + 30 \\ x^2 + x - 2 & = 2x^2 - 17x + 30 \\ 0 & = 2x^2 - x^2 - 17x - x + 30 + 2 \\ 0 & = x^2 - 18x + 32 \\ 0 & = (x - 2)(x - 16) \end{align} \begin{align} x - 2 & = 0 && \text{ or } & x - 16 & = 0 \\ x & = 2 &&& x & = 16 \end{align}
(c)
\begin{align} {2x + 3 \over 3x + 5} & = {x - 1 \over 3x - 1} \\ {(2x + 3)(3x - 1) \over (3x + 5)(3x - 1)} & = {(x - 1)(3x + 5) \over (3x + 5)(3x - 1)} \\ \\ (2x + 3)(3x - 1) & = (x - 1)(3x + 5) \\ 6x^2 - 2x + 9x - 3 & = 3x^2 + 5x - 3x - 5 \\ 6x^2 + 7x - 3 & = 3x^2 + 2x - 5 \\ 6x^2 - 3x^2 + 7x - 2x - 3 + 5 & = 0 \\ 3x^2 + 5x + 2 & = 0 \\ (3x + 2)(x + 1) & = 0 \end{align} \begin{align} 3x + 2 & = 0 && \text{ or } & x + 1 & = 0 \\ 3x & = -2 &&& x & = -1 \\ x & = -{2 \over 3} \end{align}
(d)
\begin{align} {(x - 2)(x - 3) \over (x - 1)(x - 5)} & = {5 \over 3} \\ {3(x - 2)(x - 3) \over 3(x - 1)(x - 5)} & = {5(x - 1)(x - 5) \over 3(x - 1)(x - 5)} \\ \\ 3(x - 2)(x - 3) & = 5(x - 1)(x - 5) \\ 3(x^2 - 3x - 2x + 6) & = 5(x^2 - 5x - x + 5) \\ 3(x^2 - 5x + 6) & = 5(x^2 - 6x + 5) \\ 3x^2 - 15x + 18 & = 5x^2 - 30x + 25 \\ 0 & = 5x^2 - 3x^2 - 30x + 15x + 25 - 18 \\ 0 & = 2x^2 - 15x + 7 \\ 0 & = (2x - 1)(x - 7) \end{align} \begin{align} 2x - 1 & = 0 && \text{ or } & x - 7 & = 0 \\ 2x & = 1 &&& x & = 7 \\ x & = {1 \over 2} \end{align}
Question 2
(a)
\begin{align} {x \over 1} + {1 \over x} & = {2 \over 1} \\ {x^2 \over x} + {1 \over x} & = {2x \over x} \\ {x^2 + 1 \over x} & = {2x \over x} \\ \\ x^2 + 1 & = 2x \\ x^2 - 2x + 1 & = 0 \\ (x - 1)(x - 1) & = 0 \\ (x - 1)^2 & = 0 \\ x - 1 & = 0 \\ x & = 1 \end{align}
(b)
\begin{align} {x \over 2} + {1 \over 1} & = {24 \over x} \\ {x^2 \over 2x} + {2x \over 2x} & = {2(24) \over 2x} \\ {x^2 + 2x \over 2x} & = {48 \over 2x} \\ \\ x^2 + 2x & = 48 \\ x^2 + 2x - 48 & = 0 \\ (x + 8)(x - 6) & = 0 \end{align} \begin{align} x + 8 & = 0 && \text{ or } & x - 6 & = 0 \\ x & = -8 &&& x & = 6 \end{align}
(c)
\begin{align} {x - 2 \over 5} & = {1 \over 1} - {x - 1 \over 2x - 3} \\ {(x - 2)(2x - 3) \over 5(2x - 3)} & = {5(2x - 3) \over 5(2x - 3)} - {5(x - 1) \over 5(2x - 3)} \\ {(x - 2)(2x - 3) \over 5(2x - 3)} & = {5(2x - 3) - 5(x - 1) \over 5(2x - 3)} \\ \\ (x - 2)(2x - 3) & = 5(2x - 3) - 5(x - 1) \\ 2x^2 - 3x - 4x + 6 & = 10x - 15 - 5x + 5 \\ 2x^2 - 7x + 6 & = 5x - 10 \\ 2x^2 - 7x - 5x + 6 + 10 & = 0 \\ 2x^2 - 12x + 16 & = 0 \\ x^2 - 6x + 8 & = 0 \\ (x - 2)(x - 4) & = 0 \end{align} \begin{align} x - 2 & = 0 && \text{ or } & x - 4 & = 0 \\ x & = 2 &&& x & = 4 \end{align}
(d)
\begin{align} {7 - x \over x + 2} - {1 \over x - 1} & = {1 \over 3} \\ {3(7 - x)(x - 1) \over 3(x + 2)(x - 1)} - {3(x + 2) \over 3(x + 2)(x - 1)} & = {(x + 2)(x - 1) \over 3(x + 2)(x - 1)} \\ {3(7 - x)(x - 1) - 3(x + 2) \over 3(x + 2)(x - 1)} & = {(x + 2)(x - 1) \over 3(x + 2)(x - 1)} \\ \\ 3(7 - x)(x - 1) - 3(x + 2) & = (x + 2)(x - 1) \\ 3(7x - 7 - x^2 + x) - 3x - 6 & = x^2 - x + 2x - 2 \\ 3(- x^2 + 8x - 7) - 3x - 6 & = x^2 + x - 2 \\ - 3x^2 + 24x - 21 - 3x - 6 & = x^2 + x - 2 \\ - 3x^2 + 21x - 27 & = x^2 + x - 2 \\ 0 & = x^2 + 3x^2 + x - 21x - 2 + 27 \\ 0 & = 4x^2 - 20x + 25 \\ 0 & = (2x - 5)(2x - 5) \\ 0 & = (2x - 5)^2 \\ 0 & = 2x - 5 \\ 5 & = 2x \\ {5 \over 2} & = x \end{align}
Question 3
(a)
\begin{align} {3 \over x - 4} + {5x + 2 \over x^2 - 16} & = {3 \over x - 4} + {5x + 2 \over x^2 - 4^2} \\ & = {3 \over x - 4} + {5x + 2 \over (x + 4)(x - 4)} \phantom{000000} [a^2 - b^2 = (a + b)(a - b)] \\ & = {3(x + 4) \over (x + 4)(x - 4)} + {5x + 2 \over (x + 4)(x - 4)} \\ & = {3(x + 4) + 5x + 2 \over (x + 4)(x - 4)} \\ & = {3x + 12 + 5x + 2 \over (x + 4)(x - 4)} \\ & = {8x + 14 \over (x + 4)(x - 4)} \\ & = {2(4x + 7) \over (x + 4)(x - 4)} \end{align}
(b)
\begin{align} {3 \over x - 4} + {5x + 2 \over x^2 - 16} & = 6 \\ {8x + 14 \over (x + 4)(x - 4)} & = {6 \over 1} \\ {8x + 14 \over (x + 4)(x - 4)} & = {6(x + 4)(x - 4) \over (x + 4)(x - 4)} \\ \\ 8x + 14 & = 6(x + 4)(x - 4) \\ 8x + 14 & = 6(x^2 - 16) \\ 8x + 14 & = 6x^2 - 96 \\ 0 & = 6x^2 - 8x - 96 - 14 \\ 0 & = 6x^2 - 8x - 110 \\ 0 & = 3x^2 - 4x - 55 \\ 0 & = (3x + 11)(x - 5) \end{align} \begin{align} 3x + 11 & = 0 && \text{ or } & x - 5 & = 0 \\ 3x & = -11 &&& x & = 5 \\ x & = -{11 \over 3} \end{align}
Question 4
(a)
\begin{align} {5 \over 3x - 1} + {5 \over 9x^2 - 6x + 1} & = {5 \over 3x - 1} + {5 \over (3x - 1)(3x - 1)} \\ & = {5(3x - 1) \over (3x - 1)^2} + {5 \over (3x - 1)^2} \\ & = {5(3x - 1) + 5 \over (3x - 1)^2} \\ & = {15x - 5 + 5 \over (3x - 1)^2} \\ & = {15x \over (3x - 1)^2} \end{align}
(b)
\begin{align} {5 \over 3x - 1} + {5 \over 9x^2 - 6x + 1} & = -1 \\ {15x \over (3x - 1)^2} & = {- (3x - 1)^2 \over (3x - 1)^2 } \\ \\ 15x & = -(3x - 1)^2 \\ 15x & = -(9x^2 - 6x + 1) \\ 15x & = - 9x^2 + 6x - 1 \\ 0 & = - 9x^2 + 6x - 15x - 1 \\ 0 & = - 9x^2 - 9x - 1 \\ 0 & = 9x^2 + 9x + 1 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-9 \pm \sqrt{(9)^2 - 4(9)(1)} \over 2(9)} \\ & = {-9 \pm \sqrt{45} \over 18} \\ & = -0.12732 \text{ or } -0.87268 \\ & \approx -0.127 \text{ or } -0.873 \end{align}
Question 5
(a)
\begin{align} {x \over x - 2} + {x + 7 \over x + 2} & = {5 \over 1} \\ {x(x + 2) \over (x - 2)(x + 2)} + {(x + 7)(x - 2) \over (x - 2)(x + 2)} & = {5(x - 2)(x + 2) \over (x - 2)(x + 2)} \\ {x(x + 2) + (x + 7)(x - 2) \over (x - 2)(x + 2)} & = {5(x - 2)(x + 2) \over (x - 2)(x + 2)} \\ \\ x (x + 2) + (x + 7)(x - 2) & = 5(x - 2)(x + 2) \\ x^2 + 2x + x^2 - 2x + 7x - 14 & = 5( \underbrace{x^2 - 2^2}_{(a - b)(a + b) = a^2 - b^2}) \\ 2x^2 + 7x - 14 & = 5(x^2 - 4) \\ 2x^2 + 7x - 14 & = 5x^2 - 20 \\ 0 & = 5x^2 - 2x^2 - 7x - 20 + 14 \\ 0 & = 3x^2 - 7x - 6 \\ 0 & = (3x + 2)(x - 3) \end{align} \begin{align} 3x + 2 & = 0 && \text{ or } & x - 3 & = 0 \\ 3x & = -2 &&& x & = 3 \\ x & = -{2 \over 3} \end{align}
(b)
\begin{align} {3 \over x - 1} + {2x + 10 \over x^2 + 2x - 3} & = {1 \over 3} \\ {3 \over x - 1} + {2x + 10 \over (x - 1)(x + 3) } & = {1 \over 3} \\ {3(3)(x + 3) \over 3(x - 1)(x + 3)} + {3(2x + 10) \over 3(x - 1)(x + 3)} & = {(x - 1)(x + 3) \over 3(x - 1)(x + 3)} \\ {9(x + 3) + 3(2x + 10) \over 3(x - 1)(x + 3)} & = {(x - 1)(x + 3) \over 3(x - 1)(x + 3)} \\ \\ 9(x + 3) + 3(2x + 10) & = (x - 1)(x + 3) \\ 9x + 27 + 6x + 30 & = x^2 + 3x - x - 3 \\ 15x + 57 & = x^2 + 2x - 3 \\ 0 & = x^2 + 2x - 15x - 3 - 57 \\ 0 & = x^2 - 13x - 60 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-13) \pm \sqrt{(-13)^2 - 4(1)(-60)} \over 2(1)} \\ & = {13 \pm \sqrt{409} \over 2} \\ & = 16.611 \text{ or } -3.6118 \\ & \approx 16.6 \text{ or } -3.61 \end{align}
(c)
\begin{align} {4x - 3 \over x - 5} & = {2 \over 1} - {x - 5 \over x - 7} \\ {(4x - 3)(x - 7) \over (x - 5)(x - 7)} & = {2(x - 5)(x - 7) \over (x - 5)(x - 7)} - {(x - 5)^2 \over (x - 5)(x - 7)} \\ {(4x - 3)(x - 7) \over (x - 5)(x - 7)} & = {2(x - 5)(x - 7) - (x - 5)^2 \over (x - 5)(x - 7)} \\ \\ (4x - 3)(x - 7) & = 2(x - 5)(x - 7) - (x - 5)^2 \\ 4x^2 - 28x - 3x + 21 & = 2(x^2 - 7x - 5x + 35) - [ \underbrace{(x)^2 - 2(x)(5) + (5)^2}_{(a - b)^2 = a^2 - 2ab + b^2} ] \\ 4x^2 - 31x + 21 & = 2(x^2 - 12x + 35) - (x^2 - 10x + 25) \\ 4x^2 - 31x + 21 & = 2x^2 - 24x + 70 - x^2 + 10x - 25 \\ 4x^2 - 31x + 21 & = x^2 - 14x + 45 \\ 4x^2 - x^2 - 31x + 14x + 21 - 45 & = 0 \\ 3x^2 - 17x - 24 & = 0 \\ \\ x & = { -b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-17) \pm \sqrt{(-17)^2 - 4(3)(-24)} \over 2(3)} \\ & = {17 \pm \sqrt{577} \over 6} \\ & = 6.8368 \text{ or } -1.1701 \\ & \approx 6.84 \text{ or } -1.17 \end{align}
(d)
\begin{align} {7 \over x + 3} - {4 \over (x - 3)} & = {5 \over 1} \\ {7(x - 3) \over (x + 3)(x - 3)} - {4(x + 3) \over (x + 3)(x - 3)} & = {5(x + 3)(x - 3) \over (x + 3)(x - 3)} \\ {7(x - 3) - 4(x + 3) \over (x + 3)(x - 3)} & = {5(x + 3)(x - 3) \over (x + 3)(x - 3)} \\ \\ 7(x - 3) - 4(x + 3) & = 5(x + 3)(x - 3) \\ 7x - 21 - 4x - 12 & = 5 ( \underbrace{x^2 - 3^2}_{(a + b)(a - b) = a^2 - b^2} ) \\ 3x - 33 & = 5 (x^2 - 9) \\ 3x - 33 & = 5x^2 - 45 \\ 0 & = 5x^2 - 3x - 45 + 33 \\ 0 & = 5x^2 - 3x - 12 \\ \\ x & = { -b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-3) \pm \sqrt{(-3)^2 - 4(5)(-12)} \over 2(5)} \\ & = {3 \pm \sqrt{249} \over 10} \\ & = 1.8779 \text{ or } -1.2779 \\ & \approx 1.88 \text{ or } - 1.28 \end{align}
(e)
\begin{align} {1 \over x} + {2 \over x - 1} + {3 \over x + 1} & = 0 \\ {(x + 1)(x - 1) \over x(x + 1)(x - 1)} + {2x(x + 1) \over x(x + 1)(x - 1)} + {3x(x - 1) \over x(x + 1)(x - 1)} & = 0 \\ {(x + 1)(x - 1) + 2x(x + 1) + 3x(x - 1) \over x(x + 1)(x - 1)} & = {0 \over x(x + 1)(x - 1)} \\ \\ (x + 1)(x - 1) + 2x(x + 1) + 3x(x - 1) & = 0 \\ \underbrace{ (x)^2 - (1)^2 }_{ (a + b)(a - b) = a^2 - b^2 } + 2x^2 + 2x + 3x^2 - 3x & = 0 \\ x^2 - 1^2 + 5x^2 - x & = 0 \\ x^2 - 1 + 5x^2 - x & = 0 \\ 6x^2 - x - 1 & = 0 \\ (3x + 1)(2x - 1) & = 0 \end{align} \begin{align} 3x + 1 & = 0 && \text{ or } & 2x - 1 & = 0 \\ 3x & = -1 &&& 2x & = 1 \\ x & = -{1 \over 3} &&& x & = {1 \over 2} \end{align}
(f)
\begin{align} {5 \over x - 15} - {2 \over x + 6} & = {3 \over x - 9} \\ {5(x + 6)(x - 9) \over (x - 15)(x + 6)(x - 9)} - {2 (x - 15)(x - 9) \over (x - 15)(x + 6)(x - 9)} & = {3(x - 15)(x + 6) \over (x - 15)(x + 6)(x - 9)} \\ {5(x + 6)(x - 9) - 2(x - 15)(x - 9) \over (x - 15)(x + 6)(x - 9)} & = {3(x - 15)(x + 6) \over (x - 15)(x + 6)(x - 9)} \\ \\ 5(x + 6)(x - 9) - 2(x - 15)(x - 9) & = 3(x - 15)(x + 6) \\ 5(x^2 - 9x + 6x - 54) - 2(x^2 - 9x - 15x + 135) & = 3(x^2 + 6x - 15x - 90) \\ 5(x^2 - 3x - 54) - 2(x^2 - 24x + 135) & = 3(x^2 - 9x - 90) \\ 5x^2 - 15x - 270 - 2x^2 + 48x - 270 & = 3x^2 - 27x - 270 \\ 3x^2 + 33x - 540 & = 3x^2 - 27x - 270 \\ 3x^2 - 3x^2 + 33x + 27x & = -270 + 540 \\ 60x & = 270 \\ x & = {270 \over 60} \\ x & = {9 \over 2} \end{align}
Question 6
(a)
\begin{align} {x \over 1} + {1 \over x + 3} & = {4 \over 1} \\ {x(x + 3) \over x + 3} + {1 \over x + 3} & = {4(x + 3) \over x + 3} \\ {x(x + 3) + 1 \over x + 3} & = {4(x + 3) \over x + 3} \\ \\ x(x + 3) + 1 & = 4(x + 3) \\ x^2 + 3x + 1 & = 4x + 12 \\ x^2 + 3x - 4x + 1 - 12 & = 0 \\ x^2 - x - 11 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-1) \pm \sqrt{(-1)^2 - 4(1)(-11)} \over 2(1)} \\ & = {1 \pm \sqrt{45} \over 2} \\ & = 3.8541 \text{ or } -2.8541 \\ & \approx 3.85 \text{ or } -2.85 \end{align}
(b)
\begin{align} {3 \over x - 2} + {5 \over 1} & = {1 \over x + 1} \\ {3(x + 1) \over (x - 2)(x + 1)} + {5(x - 2)(x + 1) \over (x - 2)(x + 1)} & = {x - 2 \over (x - 2)(x + 1)} \\ {3(x + 1) + 5(x - 2)(x + 1) \over (x - 2)(x + 1)} & = {x - 2 \over (x - 2)(x + 1)} \\ \\ 3(x + 1) + 5(x - 2)(x + 1) & = x - 2 \\ 3x + 3 + 5(x^2 + x - 2x - 2) & = x - 2 \\ 3x + 3 + 5(x^2 - x - 2) & = x - 2 \\ 3x + 3 + 5x^2 - 5x - 10 & = x - 2 \\ 5x^2 - 2x - 7 & = x - 2 \\ 5x^2 - 2x - x - 7 + 2 & = 0 \\ 5x^2 - 3x - 5 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-3) \pm \sqrt{(-3)^2 - 4(5)(-5)} \over 2(5)} \\ & = {3 \pm \sqrt{109} \over 10} \\ & = 1.344 \text{ or } -0.74403 \\ & \approx 1.34 \text{ or } -0.744 \end{align}
(c)
\begin{align} {4 \over x - 3} + {1 \over x + 3} & = {6 \over 1} \\ {4(x + 3) \over (x - 3)(x + 3)} + {x - 3 \over (x - 3)(x + 3)} & = {6(x - 3)(x + 3) \over (x - 3)(x + 3)} \\ {4(x + 3) + x - 3 \over (x - 3)(x + 3)} & = {6(x - 3)(x + 3) \over (x - 3)(x + 3)} \\ \\ 4(x + 3) + x - 3 & = 6(x - 3)(x + 3) \\ 4x + 12 + x - 3 & = 6(x^2 - 3^2) \phantom{000000} [(a - b)(a + b) = a^2 - b^2] \\ 5x + 9 & = 6(x^2 - 9) \\ 5x + 9 & = 6x^2 - 54 \\ 0 & = 6x^2 - 5x - 54 - 9 \\ 0 & = 6x^2 - 5x - 63 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-5) \pm \sqrt{(-5)^2 - 4(6)(-63)} \over 2(6)} \\ & = {5 \pm \sqrt{1537} \over 12} \\ & = 3.6837 \text{ or } -2.8504 \\ & \approx 3.68 \text{ or } -2.85 \end{align}
(d)
\begin{align} {7 \over x^2 - x - 6} - {x \over x + 2} & = 0 \\ {7 \over (x + 2)(x - 3)} & = {x \over x + 2} \\ {7 \over (x + 2)(x - 3)} & = {x(x - 3) \over (x + 2)(x - 3)} \\ \\ 7 & = x (x - 3) \\ 7 & = x^2 - 3x \\ 0 & = x^2 - 3x - 7 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-3) \pm \sqrt{(-3)^2 - 4(1)(-7)} \over 2(1)} \\ & = {3 \pm \sqrt{37} \over 2} \\ & = 4.5413 \text{ or } -1.5413 \\ & \approx 4.54 \text{ or } -1.54 \end{align}
Question 7
(a)
\begin{align} {k \over x + 2} + {k + 1 \over x - 3} & = 2 \\ \\ \text{Let } x = -1, & \\ {k \over -1 + 2} + {k + 1 \over -1 - 3} & = 2 \\ {k \over 1} + {k + 1 \over - 4} & = 2 \\ {4k \over 4} - {k + 1 \over 4} & = {8 \over 4} \\ {4k - (k + 1) \over 4} & = {8 \over 4} \\ \\ 4k - (k + 1) & = 8 \\ 4k - k - 1 & = 8 \\ 3k - 1 & = 8 \\ 3k & = 8 + 1 \\ 3k & = 9 \\ k & = {9 \over 3} \\ k & = 3 \end{align}
(b)
\begin{align}
{k \over x + 2} + {k + 1 \over x - 3} & = 2 \\
{3 \over x + 2} + {3 + 1 \over x - 3} & = {2 \over 1} \\
{3(x - 3) \over (x + 2)(x - 3)} + {4(x + 2) \over (x + 2)(x - 3)} & = {2(x + 2)(x - 3) \over (x + 2)(x - 3)} \\
{3(x - 3) + 4(x + 2) \over (x + 2)(x - 3)} & = {2(x + 2)(x - 3) \over (x + 2)(x - 3)} \\
\\
3(x - 3) + 4(x + 2) & = 2(x + 2)(x - 3) \\
3x - 9 + 4x + 8 & = 2(x^2 - 3x + 2x - 6) \\
7x - 1 & = 2(x^2 - x - 6) \\
7x - 1 & = 2x^2 - 2x - 12 \\
0 & = 2x^2 - 2x - 7x - 12 + 1 \\
0 & = 2x^2 - 9x - 11 \\
0 & = (x + 1)(2x - 11)
\end{align}
\begin{align}
x + 1 & = 0
&& \text{ or } &
2x - 11 & = 0 \\
x & = -1
&&&
2x & = 11 \\
& &&&
x & = {11 \over 2}
\end{align}
$$ \text{Other solution is } x = {11 \over 2} $$
Question 8
(a)
$$ \text{Equations 1 and 2 since } {5 \over 0} \text{ and } {10 \over 2(0)} \text{ are undefined} $$
(b)
\begin{align} x = 2 \text{ makes the } & \text{denominators zero,} \\ \text{so it is not a } & \text{valid solution} \\ \\ \therefore \text{Equation (1) has } & \text{no real solution} \end{align}
Question 9
(a)(i)
\begin{align} {1 \over x} - {1 \over x + 1} & = {x + 1 \over x(x + 1)} - {x \over x(x + 1)} \\ & = {x + 1 - x \over x(x + 1)} \\ & = {1 \over x(x + 1)} \\ & = {1 \over x^2 + x} \end{align}
(a)(ii)
\begin{align} {1 \over x + 1} - {1 \over x + 2} & = {x + 2 \over (x + 1)(x + 2)} - {x + 1 \over (x + 1)(x + 2)} \\ & = {x + 2 - (x + 1) \over (x + 1)(x + 2)} \\ & = {x + 2 - x - 1 \over (x + 1)(x + 2)} \\ & = {1 \over (x + 1)(x + 2)} \\ & = {1 \over x^2 + 2x + x + 2} \\ & = {1 \over x^2 + 3x + 2} \end{align}
(b)
\begin{align} {1 \over x^2 + x} + {1 \over x^2 + 3x + 2} & = 1 \\ \underbrace{ {1 \over x} - {1 \over x + 1} }_\text{(a)(i)} + \underbrace{ {1 \over x + 1} - {1 \over x + 2} }_\text{(a)(ii)} & = 1 \\ {1 \over x} - {1 \over x + 2} & = {1 \over 1} \\ {x + 2 \over x(x + 2)} - {x \over x(x + 2)} & = {x(x + 2) \over x(x + 2)} \\ {x + 2 - x \over x(x + 2)} & = {x(x + 2) \over x(x + 2)} \\ {2 \over x(x + 2)} & = {x(x + 2) \over x(x + 2)} \\ \\ 2 & = x(x + 2) \\ 2 & = x^2 + 2x \\ 0 & = x^2 + 2x - 2 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(2) \pm \sqrt{(2)^2 - 4(1)(-2)} \over 2(1)} \\ & = {-2 \pm \sqrt{12} \over 2} \\ & = 0.73205 \text{ or } -2.732 \\ & \approx 0.732 \text{ or } -2.73 \end{align}
Ex 1.5: Solving Real-World Problems Involving Quadratic Equations
Question 1
(a)
\begin{align} \text{Next positive integer} & = x + 1 \\ \\ \text{Product} & = x(x + 1) \end{align}
(b)(i)
\begin{align} x(x + 1) & = 1260 \\ x^2 + x & = 1260 \\ x^2 + x - 1260 & = 0 \phantom{00} \text{ (Shown)} \end{align}
(b)(ii)
\begin{align}
x^2 + x - 1260 & = 0 \\
(x - 35)(x + 36) & = 0
\end{align}
\begin{align}
x - 35 & = 0
&& \text{ or } &
x + 36 & = 0 \\
x & = 35
&&&
x & = -36 \text{ (Reject, } x \text{ is a positive integer})
\end{align}
$$ \text{Two integers are } 35 \text{ and } 36 $$
Question 2
(a)
\begin{align} \text{Area of triangle} & = {1 \over 2} \times \text{Base} \times \text{Height} \\ & = {1 \over 2} \times x \times (x - 3) \\ & = {1 \over 2}x (x - 3) \\ & = {1 \over 2}x^2 - {3 \over 2}x \\ \\ 90 & = {1 \over 2}x^2 - {3 \over 2}x \\ 2(90) & = 2 \left( {1 \over 2}x^2 - {3 \over 2}x \right) \\ 180 & = x^2 - 3x \\ 0 & = x^2 - 3x - 180 \phantom{00} \text{ (Shown)} \end{align}
(b)
\begin{align}
0 & = x^2 - 3x - 180 \\
0 & = (x - 15)(x + 12)
\end{align}
\begin{align}
x - 15 & = 0
&& \text{ or } &
x + 12 & = 0 \\
x & = 15
&&&
x & = -12 \text{ (Reject, base of triangle } > 0)
\end{align}
$$ \text{Base} = 15 \text{ cm} $$
Question 3
(a)
\begin{align} \text{Perimeter} & = 2l + 2b \\ 68 & = 2x + 2b \\ 68 - 2x & = 2b \\ {1 \over 2}(68 - 2x) & = b \\ 34 - x & = b \\ \\ \text{Breadth} & = (34 - x) \text{ cm} \end{align}
(b)
\begin{align} \text{Area} & = l \times b \\ 253 & = x \times (34 - x) \\ 253 & = x(34 - x) \\ 253 & = 34x - x^2 \\ 0 & = -x^2 + 34x - 253 \\ 0 & = x^2 - 34x + 253 \phantom{00} \text{ (Shown)} \end{align}
(c)
\begin{align}
0 & = x^2 - 34x + 253 \\
0 & = (x - 11)(x - 23)
\end{align}
\begin{align}
x - 11 & = 0
&& \text{ or } &
x - 23 & = 0 \\
x & = 11
&&&
x & = 23
\end{align}
$$ \text{Dimensions: } 23 \text{ cm by } 11 \text{ cm} $$
Question 4
(a)
\begin{align} \text{Length of } B & = (x + 5) \text{ cm} \end{align}
(b)
\begin{align} \text{Area of square} & = l \times l = l^2 \\ \\ \text{Area of } B & = (x + 5)^2 \text{ cm}^2 \end{align}
(c)
\begin{align} \text{Sum of area} & = \text{Area of } A + \text{Area of } B \\ 1525 & = x^2 + (x + 5)^2 \\ 1525 & = x^2 + \underbrace{ (x)^2 + 2(x)(5) + (5)^2 }_{(a + b)^2 = a^2 + 2ab + b^2} \\ 1525 & = x^2 + x^2 + 10x + 25 \\ 1525 & = 2x^2 + 10x + 25 \\ 0 & = 2x^2 + 10x + 25 - 1525 \\ 0 & = 2x^2 + 10x - 1500 \\ 0 & = x^2 + 5x - 750 \\ 0 & = (x - 25)(x + 30) \end{align} \begin{align} x - 25 & = 0 && \text{ or } & x + 30 & = 0 \\ x & = 25 &&& x & = -30 \text{ (Reject, since length of tile } A > 0) \end{align}
Question 5
(a)(i)
$$ (24 - x) \text{ cm} $$
(a)(ii)
$$ (17 + x) \text{ cm} $$
(a)(iii)
\begin{align} \text{New area} & = (24 - x)(17 + x) \text{ cm}^2 \end{align}
(b)
\begin{align} \text{New area} & = \text{Initial area} + 12 \\ (24 - x)(17 + x) & = 24 \times 17 + 12 \\ 408 + 24x - 17x - x^2 & = 420 \\ 408 + 7x - x^2 & = 420 \\ 0 & = x^2 - 7x + 420 - 408 \\ 0 & = x^2 - 7x + 12 \\ 0 & = (x - 3)(x - 4) \end{align} \begin{align} x - 3 & = 0 && \text{ or } & x - 4 & = 0 \\ x & = 3 &&& x & = 4 \end{align}
Question 6
(a)(i)
$$ 34 - x $$
(a)(ii)
\begin{align} 34 - x + 4 & = 38 - x \end{align}
(a)(iii)
$$ (x + 4)^2 $$
(b)
\begin{align}
\text{Father's age (in 4 years' time)} & = \text{Square of Mingfa's age (in 4 years' time)} \\
38 - x & = (x + 4)^2 \\
38 - x & = (x + 4)(x + 4) \\
38 - x & = x^2 + 4x + 4x + 16 \\
38 - x & = x^2 + 8x + 16 \\
0 & = x^2 + 8x + x + 16 - 38 \\
0 & = x^2 + 9x - 22 \\
0 & = (x - 2)(x + 11)
\end{align}
\begin{align}
x - 2 & = 0
&& \text{ or } &
x + 11 & = 0 \\
x & = 2
&&&
x & = -11 \text{ (Reject, as age } \ge 0 )
\end{align}
\begin{align}
\text{Mingfa's present age} & = x \\
& = 2 \text{ years}
\end{align}
Question 7
(a) Note: Hypotenuse is the longest side of the right-angled triangle (opposite the right angle)
\begin{align} \text{Hypotenuse} & = 10 - x - 2x \\ & = (10 - 3x) \text{ cm} \\ \\ \text{By Pytha}& \text{goras theorem,} \\ (10 - 3x)^2 & = x^2 + (2x)^2 \\ (10 - 3x)(10 - 3x) & = x^2 + 4x^2 \\ 100 - 30x - 30x + 9x^2 & = 5x^2 \\ 9x^2 - 60x + 100 & = 5x^2 \\ 9x^2 - 5x^2 - 60x + 100 & = 0 \\ 4x^2 - 60x + 100 & = 0 \\ x^2 - 15x + 25 & = 0 \phantom{00} \text{ (Shown)} \end{align}
(b)
\begin{align} x^2 & - 15x + 25 = 0 \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-15) \pm \sqrt{(-15)^2 - 4(1)(25)} \over 2(1)} \\ & = {15 \pm \sqrt{125} \over 2} \\ & = 13.09 \text{ (Reject, as } 10 - 3x > 0 \implies x < {10 \over 3} \text{)} \text{ or } 1.9098 \\ \\ \text{Longest side} & = \text{Hypotenuse} \\ & = 10 - 3(1.9098) \\ & = 4.2706 \\ & \approx 4.27 \text{ cm} \end{align}
Question 8
\begin{align}
\text{Let } x \text{ represent the } & \text{smallest even number} \\
\\
\text{Next } \textbf{even} \text{ number} & = x + 2 \\
\\
\text{Last even number} & = x + 2 + 2 \\
& = x + 4 \\
\\
\text{Sum of squares} & = x^2 + (x + 2)^2 + (x + 4)^2 \\
2360 & = x^2 + (x + 2)(x + 2) + (x + 4)(x + 4) \\
2360 & = x^2 + x^2 + 2x + 2x + 4 + x^2 + 4x + 4x + 16 \\
2360 & = 3x^2 + 12x + 20 \\
0 & = 3x^2 + 12x + 20 - 2360 \\
0 & = 3x^2 + 12x - 2340 \\
0 & = x^2 + 4x - 780 \\
0 & = (x - 26)(x + 30)
\end{align}
\begin{align}
x - 26 & = 0
&& \text{ or } &
x + 30 & = 0 \\
x & = 26
&&&
x & = -30 \text{ (Reject, } x \text{ represents a positive even number})
\end{align}
\begin{align}
\text{Smallest even number} & = x \\
& = 26
\end{align}
Question 9
\begin{align} \text{Length of typing area} & = (29 - 2x) \text{ cm} \\ \\ \text{Breadth of typing area} & = (21 - 2x) \text{ cm} \\ \\ \text{Area of typing area} & = (29 - 2x)(21 - 2x) \\ 380 & = 609 - 58x - 42x + 4x^2 \\ 380 & = 609 - 100x + 4x^2 \\ 0 & = 4x^2 - 100x + 609 - 380 \\ 0 & = 4x^2 - 100x + 229 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-100) \pm \sqrt{(-100)^2 - 4(4)(229)} \over 2(4)} \\ & = {100 \pm \sqrt{ 6336 } \over 8} \\ & = 22.449 \text{ (Reject, since width of paper is 21 cm}) \text{ or } 2.5501 \\ \\ \text{Width of margin} & = x \\ & = 2.5501 \\ & \approx 2.6 \text{ cm (1 d.p.)} \end{align}
Question 10
\begin{align}
\text{Let } x \text{ represent the price of} & \text{ a keyboard} \\
\\
\text{No. of keyboards purchased} & = {540 \over x} \\
\\
\text{No. of keyboards purchased (w. reduction)} & = {540 \over x - 2} \\
\\
{540 \over x - 2} - {540 \over x} & = 3 \\
{540x \over x(x - 2)} - {540(x - 2) \over x(x - 2)} & = {3 \over 1} \\
{540x - 540(x - 2) \over x(x - 2)} & = {3x(x - 2) \over x(x - 2)} \\
\\
540x - 540(x - 2) & = 3x(x - 2) \\
540x - 540x + 1080 & = 3x^2 - 6x \\
1080 & = 3x^2 - 6x \\
0 & = 3x^2 - 6x - 1080 \\
0 & = x^2 - 2x - 360 \\
0 & = (x - 20) (x + 18)
\end{align}
\begin{align}
x - 20 & = 0
&& \text{ or } &
x + 18 & = 0 \\
x & = 20
&&&
x & = -18 \text{ (Reject, as price of keyboard } > 0)
\end{align}
$$ \text{Price of one keyboard} = \$ 20 $$
Question 11
(a)
\begin{align} \text{Area} & = x \times 3 + (2x + 1 - x) \times (3 - x) \\ & = 3x + (x + 1)(3 - x) \\ & = 3x + (3x - x^2 + 3 - x) \\ & = 3x + 3x - x^2 + 3 - x \\ & = - x^2 + 5x + 3 \phantom{00} \text{ (Shown)} \end{align}
(b)
\begin{align} 8.25 & = - x^2 + 5x + 3 \\ 0 & = -x^2 + 5x + 3 - 8.25 \\ 0 & = - x^2 + 5x - 5.25 \\ 0 & = x^2 - 5x + 5.25 \\ 4(0) & = 4(x^2 - 5x + 5.25) \\ 0 & = 4x^2 - 20x + 21 \\ 0 & = (2x - 3)(2x - 7) \end{align} \begin{align} 2x - 3 & = 0 && \text{ or } & 2x - 7 & = 0 \\ 2x & = 3 &&& 2x & = 7 \\ x & = {3 \over 2} = 1.5 &&& x & = {7 \over 2} = 3.5 \text{ (Reject, since } x < 3) \end{align}
Question 12
\begin{align} \text{Let } x \text{ represent the } & \text{length of the square cardboard} \\ \\ \text{Length of base of cuboid} & = x - 6 - 6 \\ & = (x - 12) \text{ cm} \\ \\ \text{Height of cuboid} & = 6 \text{ cm} \\ \\ \text{Volume} & = (x - 12) \times (x - 12) \times 6 \\ 2000 & = 6(x - 12)(x - 12) \\ 2000 & = 6(x^2 - 12x - 12x + 144) \\ 2000 & = 6(x^2 - 24x + 144) \\ 2000 & = 6x^2 - 144x + 864 \\ 0 & = 6x^2 - 144x + 864 - 2000 \\ 0 & = 6x^2 - 144x - 1136 \\ 0 & = 3x^2 - 72x - 568 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-72) \pm \sqrt{(-72)^2 - 4(3)(-568)} \over 2(3)} \\ & = {72 \pm \sqrt{ 12 \phantom{.} 000} \over 6} \\ & = 30.257 \text{ or } -6.2574 \text{ (Reject, as } x > 12) \\ \\ \text{Length} & = x \\ & = 30.257 \\ & \approx 30.3 \text{ cm} \end{align}
Question 13
(a)
\begin{align} \text{Area} & = {1 \over 2} \times b \times h \\ & = {1 \over 2} \times 4 \times x \\ & = 2x \text{ cm}^2 \end{align}
(b)
\begin{align} \text{Area of paper} & = 4x \times (5x - 1) \\ & = 4x(5x - 1) \\ & = 20x^2 - 4x \\ \\ \text{Area of rectangle } DEFG & = 4 \times x \\ & = 4x \\ \\ \text{Area of shaded region} & = 20x^2 - 4x - 4x \\ & = (20x^2 - 8x) \text{ cm}^2 \phantom{00} \text{ (Shown)} \end{align}
(c)(i)
\begin{align} 450 & = 20x^2 - 8x \\ 0 & = 20x^2 - 8x - 450 \\ 0 & = 10x^2 - 4x - 225 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-4) \pm \sqrt{(-4)^2 - 4(10)(-225)} \over 2(10)} \\ & = {4 \pm \sqrt{9016} \over 20} \\ & = 4.9476 \text{ or } -4.5476 \text{ (Reject, as } x > 0) \\ & \approx 4.95 \end{align}
(c)(ii)
\begin{align} \text{By Pytha} & \text{goras theorem,} \\ GE^2 & = GD^2 + DE^2 \\ GE^2 & = (4)^2 + (x)^2 \\ GE^2 & = 16 + x^2 \\ GE^2 & = 16 + (4.9476)^2 \\ GE^2 & = 40.479 \\ GE & = \sqrt{40.479} \\ & = 6.3623 \text{ cm} \\ \\ \text{Perimeter} & = GE + EC + CB + BA + AG \\ & = 6.3623 + [5(4.9476) - 1 - 4.9476] + 4(4.9476) + [5(4.9476) - 1] + [4(4.9476) - 4] \\ & = 84.472 \\ & \approx 84.5 \text{ cm} \end{align}
Question 14
\begin{align} \text{Let } x \text{ represent the time (in hours } & \text{from noon) when they are 10 km apart} \\ \\ \text{Distance} & = \text{Speed} \times \text{Time} \\ \\ \text{Distance travelled by Ada} & = 4 \times x \\ & = 4x \text{ km} \\ \\ \text{Distance travelled by Bob} & = 4 \times (x - 1) \\ & = 4(x - 1) \\ & = (4x - 4) \text{ km} \\ \\ \text{By Pytha} & \text{goras theorem,} \\ (\text{Distance between Ada and Bob})^2 & = (4x)^2 + (4x - 4)^2 \\ 10^2 & = 16x^2 + (4x - 4)(4x - 4) \\ 100 & = 16x^2 + 16x^2 - 16x - 16x + 16 \\ 100 & = 32x^2 - 32x + 16 \\ 0 & = 32x^2 - 32x + 16 - 100 \\ 0 & = 32x^2 - 32x - 84 \\ 0 & = 8x^2 - 8x - 21 \\ \\ x & = {-b \pm \sqrt{b^2 -4ac} \over 2a} \\ & = {-(-8) \pm \sqrt{(-8)^2 - 4(8)(-21)} \over 2(8)} \\ & = {8 \pm \sqrt{736} \over 16} \\ & = 2.1955 \text{ or } -1.1955 \text{ (Reject, as } x > 0) \\ \\ 2.1955 \text{ h} & = 2 \text{ h } (0.1955 \times 60) \text{ mins} \\ & \approx 2 \text{ h } 12 \text{ mins} \\ \\ \\ \therefore \text{They would } & \text{be 10 km apart at 2.12 pm} \end{align}
Question 15
\begin{align}
\text{Let } x \text{ represent the } & \text{numerator of the fraction} \\
\\
\text{Fraction} & = {x \over x + 2} \\
\\
\text{Increased fraction} & = {x + 3 \over x + 2 + 3} \\
& = {x + 3 \over x + 5} \\
\\
{x + 3 \over x + 5} - {x \over x + 2} & = {1 \over 18} \\
{(x + 3)(x + 2) \over (x + 5)(x + 2)} - {x(x + 5) \over (x + 5)(x + 2)} & = {1 \over 18} \\
{(x + 3)(x + 2) - x(x + 5) \over (x + 5)(x + 2)} & = {1 \over 18} \\
{x^2 + 2x + 3x + 6 - x^2 - 5x \over (x + 5)(x + 2)} & = {1 \over 18} \\
{6 \over (x + 5)(x + 2)} & = {1 \over 18} \\
{18(6) \over 18(x + 5)(x + 2)} & = {(x + 5)(x + 2) \over 18(x + 5)(x + 2)} \\
\\
18(6) & = (x + 5)(x + 2) \\
108 & = x^2 + 2x + 5x + 10 \\
108 & = x^2 + 7x + 10 \\
0 & = x^2 + 7x + 10 - 108 \\
0 & = x^2 + 7x - 98 \\
0 & = (x - 7)(x + 14)
\end{align}
\begin{align}
x - 7 & = 0
&& \text{ or } &
x + 14 & = 0 \\
x & = 7
&&&
x & = -14 \text{ (Reject, since } x > 0)
\end{align}
$$ \text{Fraction} = {x \over x + 2} = {7 \over 7 + 2} = {7 \over 9} $$
Question 16
(a)(i)
$$ 80 - 4 - 4 = 72 $$
(a)(ii)
\begin{align} \text{Sales} & = 70 \times 72 \\ & = \$ 5040 \end{align}
(b)(i)
$$ 80 - 4n $$
(b)(ii)
\begin{align} \text{Sales} & = (60 + 5n)(80 - 4n) \\ & = 4800 - 240n + 400n - 20n^2 \\ & = \$ (4800 + 160n - 20n^2) \end{align}
(b)(iii)
\begin{align} 5100 & = 4800 + 160n - 20n^2 \\ 0 & = 4800 - 5100 + 160n - 20n^2 \\ 0 & = -300 + 160n - 20n^2 \\ 0 & = -20n^2 + 160n - 300 \\ 0 & = 20n^2 - 160n + 300 \\ 0 & = n^2 - 8n + 15 \\ 0 & = (n - 3)(n - 5) \end{align} \begin{align} n - 3 & = 0 && \text{ or } & n - 5 & = 0 \\ n & = 3 &&& n & = 5 \\ \\ \text{Price per copy} & = 60 + 5(3) &&& \text{Price per copy} & = 60 + 5(5) \\ & = \$75 &&& & = \$ 85 \end{align}
(b)(iv)
\begin{align} \text{By symmetry, maximum is } & \text{midway between } n = 3 \text{ and } n = 5 \\ n & = {3 + 5 \over 2} = 4 \\ \\ \text{Price} & = 60 + 5(4) \\ & = \$ 80 \end{align}
Question 17
(a)
| t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| h | 0 | 33 | 56 | 69 | 72 | 65 | 48 | 21 | -16 |
(b) From the graph, the dolphin was above sea level from t = 0 to t = 7.6
$$ \text{Duration} = 7.6 \text{ s} $$
(c)
\begin{align} & T = 7.6 \\ \\ & \text{After } t = 7.6, \text{ the dolphin is underwater} \end{align}
Question 18
(a)
\begin{align} y & = C_1 - C_2 \\ y & = 2x^2 + 7x + 3 - (x^2 + 10x + 11) \\ y & = 2x^2 + 7x + 3 - x^2 - 10x - 11 \\ y & = x^2 - 3x - 8 \end{align}
(b)
| x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| y | -8 | -10 | -10 | -8 | -4 | 2 | 10 | 20 | 32 |
(c)
\begin{align} \text{For cost to be the same, } C_1 & = C_2 \\ C_1 - C_2 & = 0 \\ y & = 0 \\ \\ \text{From graph, } & x = 4.7 \end{align}
Ex 1.6: Sketching the Graphs of Quadratic Functions
Question 1
(a)
\begin{align} y & = (x - 1)(x - 4) \\ y & = x^2 - 4x - x + 4 \\ y & = x^2 - 5x + 4 \phantom{0000000} [\text{Minimum curve } \cup] \\ \\ \text{Let } & x = 0, \\ y & = (0 - 1)(0 - 4) \\ y & = 4 \phantom{0000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = (x - 1)(x - 4) \\ \\ x - 1 & = 0 \phantom{0} \text{ or } \phantom{0} x - 4 = 0 \\ x & = 1 \phantom{00000000} x = 4 \phantom{0000} [x \text{-intercepts}] \\ \\ \text{Line of symmetry, } x & = {1 + 4 \over 2} \\ x & = 2.5 \\ \\ \text{Let } & x = 2.5, \\ y & = (2.5 - 1)(2.5 - 4) \\ y & = -2.25 \\ \\ \text{Turning point: } & (2.5, -2.25) \end{align}
(b)
\begin{align} y & = (x + 3)(x - 1) \\ y & = x^2 - x + 3x - 3 \\ y & = x^2 + 2x - 3 \phantom{0000000} [\text{Minimum curve } \cup] \\ \\ \text{Let } & x = 0, \\ y & = (0 + 3)(0 - 1) \\ y & = - 3 \phantom{0000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = (x + 3)(x - 1) \\ \\ x + 3 & = 0 \phantom{0} \text{ or } \phantom{0} x - 1 = 0 \\ x & = -3 \phantom{000000(} x = 1 \phantom{00000} [x \text{-intercepts}] \\ \\ \text{Line of symmetry, } x & = {-3 + 1 \over 2} \\ x & = -1 \\ \\ \text{Let } & x = -1, \\ y & = (-1 + 3)(-1 - 1) \\ y & = -4 \\ \\ \text{Turning point: } & (-1, -4) \end{align}
(c)
\begin{align} y & = -(x + 2)(x + 3) \\ y & = -(x^2 + 3x + 2x + 6) \\ y & = -(x^2 + 5x + 6) \\ y & = -x^2 - 5x - 6 \phantom{0000000} [\text{Maximum curve } \cap] \\ \\ \text{Let } & x = 0, \\ y & = -(0 + 2)(0 + 3) \\ y & = -6 \phantom{0000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = -(x + 2)(x + 3) \\ \\ x + 2 & = 0 \phantom{0} \text{ or } \phantom{0} x + 3 = 0 \\ x & = -2 \phantom{000000(} x = -3 \phantom{0000} [x \text{-intercepts}] \\ \\ \text{Line of symmetry, } x & = {-2 + (-3) \over 2} \\ x & = -2.5 \\ \\ \text{Let } & x = -2.5, \\ y & = -(-2.5 + 2)(-2.5 + 3) \\ y & = 0.25 \\ \\ \text{Turning point: } & (-2.5, 0.25) \end{align}
(d)
\begin{align} y & = -x(x + 5) \\ y & = - x^2 - 5x \phantom{0000000} [\text{Maximum curve } \cap] \\ \\ \text{Let } & x = 0, \\ y & = -(0)(0 + 5) \\ y & = 0 \phantom{0000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = -x(x + 5) \\ \\ -x & = 0 \phantom{0} \text{ or } \phantom{0} x + 5 = 0 \\ x & = 0 \phantom{00000000} x = -5 \phantom{0000} [x \text{-intercepts}] \\ \\ \text{Line of symmetry, } x & = {0 + (-5) \over 2} \\ x & = -2.5 \\ \\ \text{Let } & x = -2.5, \\ y & = -(-2.5)(-2.5 + 5) \\ y & = 6.25 \\ \\ \text{Turning point: } & (-2.5, 6.25) \end{align}
Question 2
(a)
\begin{align} y & = x^2 + 2 \phantom{0000000} [\text{Minimum curve } \cup] \\ y & = (x - 0)^2 + 2 \\ \\ \text{Turning point: } & (0, 2) \\ \\ \text{Let } & x = 0, \\ y & = (0)^2 + 2 \\ y & = 2 \phantom{0000000000000000} [y \text{-intercept}] \end{align}
(b)
\begin{align} y & = (x + 1)^2 \phantom{0000000} [\text{Minimum curve } \cup] \\ y & = (x + 1)^2 + 0 \\ \\ \text{Turning point: } & (-1, 0) \\ \\ \text{Let } & x = 0, \\ y & = (0 + 1)^2 \\ y & = 1 \phantom{0000000000000000} [y \text{-intercept}] \end{align}
(c)
\begin{align} y & = (x - 3)^2 + 4 \phantom{0000000} [\text{Minimum curve } \cup] \\ \\ \text{Turning point: } & (3, 4) \\ \\ \text{Let } & x = 0, \\ y & = (0 - 3)^2 + 4 \\ y & = 13 \phantom{0000000000000000} [y \text{-intercept}] \end{align}
(d)
\begin{align} y & = (x + 2)^2 - 5 \phantom{0000000} [\text{Minimum curve } \cup] \\ \\ \text{Turning point: } & (-2, -5) \\ \\ \text{Let } & x = 0, \\ y & = (0 + 2)^2 - 5 \\ y & = -1 \phantom{0000000000000000} [y \text{-intercept}] \end{align}
(e)
\begin{align} y & = (x + 3)^2 + 2 \phantom{0000000} [\text{Minimum curve } \cup] \\ \\ \text{Turning point: } & (-3, 2) \\ \\ \text{Let } & x = 0, \\ y & = (0 + 3)^2 + 2 \\ y & = 11 \phantom{0000000000000000} [y \text{-intercept}] \end{align}
(f)
\begin{align} y & = (x - 4)^2 \phantom{0000000} [\text{Minimum curve } \cup] \\ y & = (x - 4)^2 + 0 \\ \\ \text{Turning point: } & (4, 0) \\ \\ \text{Let } & x = 0, \\ y & = (0 - 4)^2 \\ y & = 16 \phantom{0000000000000000} [y \text{-intercept}] \end{align}
Question 3
(a)
\begin{align} y & = -x^2 - 3 \phantom{0000000} [\text{Maximum curve } \cap] \\ y & = -(x - 0)^2 - 3 \\ \\ \text{Turning point: } & (0, -3) \\ \\ \text{Line of symmetry, } & x = 0 \\ \\ \text{Let } & x = 0, \\ y & = -(0)^2 - 3 \\ y & = -3 \phantom{0000000000000000} [y \text{-intercept}] \end{align}
(b)
\begin{align} y & = -(x - 4)^2 \phantom{0000000} [\text{Maximum curve } \cap] \\ y & = -(x - 4)^2 + 0 \\ \\ \text{Turning point: } & (4, 0) \\ \\ \text{Line of symmetry, } & x = 4 \\ \\ \text{Let } & x = 0, \\ y & = -(0 - 4)^2 \\ y & = -16 \phantom{0000000000000000} [y \text{-intercept}] \end{align}
(c)
\begin{align} y & = -(x + 1)^2 - 2 \phantom{0000000} [\text{Maximum curve } \cap] \\ \\ \text{Turning point: } & (-1, -2) \\ \\ \text{Line of symmetry, } x & = -1 \\ \\ \text{Let } & x = 0, \\ y & = -(0 + 1)^2 - 2 \\ y & = -3 \phantom{0000000000000000} [y \text{-intercept}] \end{align}
(d)
\begin{align} y & = -(x - 2)^2 + 3 \phantom{0000000} [\text{Maximum curve } \cap] \\ \\ \text{Turning point: } & (2, 3) \\ \\ \text{Line of symmetry, } & x = 2 \\ \\ \text{Let } & x = 0, \\ y & = - (0 - 2)^2 + 3 \\ y & = -1 \phantom{0000000000000000} [y \text{-intercept}] \end{align}
(e)
\begin{align} y & = -(x + 2)^2 \phantom{0000000} [\text{Maximum curve } \cap] \\ y & = -(x + 2)^2 + 0 \\ \\ \text{Turning point: } & (-2, 0) \\ \\ \text{Line of symmetry, } & x = -2 \\ \\ \text{Let } & x = 0, \\ y & = -(0 + 2)^2 \\ y & = -4 \phantom{0000000000000000} [y \text{-intercept}] \end{align}
(f)
\begin{align} y & = -(x + 3)^2 + 1 \phantom{0000000} [\text{Maximum curve } \cap] \\ \\ \text{Turning point: } & (-3, 1) \\ \\ \text{Line of symmetry, } & x = -3 \\ \\ \text{Let } & x = 0, \\ y & = -(0 + 3)^2 + 1 \\ y & = -8 \phantom{0000000000000000} [y \text{-intercept}] \end{align}
Question 4
(a)
\begin{align} y & = 1 - x^2 \\ y & = -x^2 + 1 \phantom{0000000} [\text{Maximum curve } \cap] \\ y & = -(x - 0)^2 + 1 \\ \\ \text{Turning point: } & (0, 1) \\ \\ \text{Let } & x = 0, \\ y & = 1 - 0^2 \\ y & = 1 \phantom{0000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = 1 - x^2 \\ x^2 & = 1 \\ x & = \pm \sqrt{1} \\ x & = \pm 1 \phantom{000000000000000} [x \text{-intercepts}] \end{align}
(b)
$$ x = 0 $$
Question 5
(a)
\begin{align} y & = x^2 + 3x - 4 \phantom{0000000} [\text{Minimum curve } \cup] \\ y & = (x + 4)(x - 1) \\ \\ \text{Let } & x = 0, \\ y & = (0 + 4)(0 - 1) \\ y & = -4 \phantom{0000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = (x + 4)(x - 1) \\ \\ x + 4 & = 0 \phantom{0} \text{ or } \phantom{0} x - 1 = 0 \\ x & = -4 \phantom{000000(} x = 1 \phantom{00000} [x \text{-intercepts}] \\ \\ \text{Line of symmetry, } x & = {-4 + 1 \over 2} \\ x & = -1.5 \\ \\ \text{Let } & x = -1.5, \\ y & = (-1.5 + 4)(-1.5 - 1) \\ y & = -6.25 \\ \\ \text{Turning point: } & (-1.5, -6.25) \end{align}
(b)
\begin{align} y & = -x^2 + 3x - 2 \phantom{0000000} [\text{Maximum curve } \cap] \\ y & = -(x^2 - 3x + 2) \\ y & = -(x - 1)(x - 2) \\ \\ \text{Let } & x = 0, \\ y & = -(0 - 1)(0 - 2) \\ y & = -2 \phantom{0000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = -(x - 1)(x - 2) \\ \\ x - 1 & = 0 \phantom{0} \text{ or } \phantom{0} x - 2 = 0 \\ x & = 1 \phantom{00000000} x = 2 \phantom{00000} [x \text{-intercepts}] \\ \\ \text{Line of symmetry, } x & = {1 + 2 \over 2} \\ x & = 1.5 \\ \\ \text{Let } & x = 1.5, \\ y & = -(1.5 - 1)(1.5 - 2) \\ y & = 0.25 \\ \\ \text{Turning point: } & (1.5, 0.25) \end{align}
(c)
\begin{align} y & = -x^2 - 2x - 1 \phantom{0000000} [\text{Maximum curve } \cap] \\ y & = -(x^2 + 2x + 1) \\ y & = -(x + 1)(x + 1) \\ y & = -(x + 1)^2 \\ \\ \text{Let } & x = 0, \\ y & = -(0 + 1)^2 \\ y & = -1 \phantom{0000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = -(x + 1)^2 \\ 0 & = x + 1 \\ -1 & = x \phantom{000000000000} [x \text{-intercept, also turning point}] \\ \\ \text{Turning point: } & (-1, 0) \end{align}
(d)
\begin{align} y & = x^2 - 4 \phantom{000000000} [\text{Minimum curve } \cup] \\ y & = x^2 - 2^2 \\ y & = (x + 2)(x - 2) \phantom{000} [a^2 - b^2 = (a + b)(a - b)] \\ \\ \text{Let } & x = 0, \\ y & = (0 + 2)(0 - 2) \\ y & = -4 \phantom{0000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = (x + 2)(x - 2) \\ \\ x + 2 & = 0 \phantom{0} \text{ or } \phantom{0} x - 2 = 0 \\ x & = -2 \phantom{000000.} x = 2 \phantom{00000} [x \text{-intercepts}] \\ \\ \text{Line of symmetry, } x & = {-2 + 2 \over 2} \\ x & = 0 \\ \\ \text{Turning point: } & (0, -4) \phantom{000000000} [\text{Same as } y \text{-intercept}] \end{align}
(e)
\begin{align} y & = -6 - 5x + x^2 \\ y & = x^2 - 5x - 6 \phantom{000000000} [\text{Minimum curve } \cup] \\ y & = (x + 1)(x - 6) \\ \\ \text{Let } & x = 0, \\ y & = (0 + 1)(0 - 6) \\ y & = -6 \phantom{0000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = (x + 1)(x - 6) \\ \\ x + 1 & = 0 \phantom{0} \text{ or } \phantom{0} x - 6 = 0 \\ x & = -1 \phantom{000000.} x = 6 \phantom{00000} [x \text{-intercepts}] \\ \\ \text{Line of symmetry, } x & = {-1 + 6 \over 2} \\ x & = 2.5 \\ \\ \text{Let } & x = 2.5, \\ y & = (2.5 + 1)(2.5 - 6) \\ y & = -12.25 \\ \\ \text{Turning point: } & (2.5, -12.25) \end{align}
(f)
\begin{align} y & = 6x - x^2 - 9 \\ y & = - x^2 + 6x - 9 \phantom{000000000} [\text{Maximum curve } \cap] \\ y & = - (x^2 - 6x + 9) \\ y & = - (x - 3)(x - 3) \\ y & = -(x - 3)^2 \\ \\ \text{Let } & x = 0, \\ y & = - (0 - 3)^2 \\ y & = - 9 \phantom{0000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = - (x - 3)^2 \\ 0 & = (x - 3)^2 \\ 0 & = x - 3 \\ 3 & = x \phantom{000000000000000000} [x \text{-intercept and turning point}] \\ \\ \text{Line of symmetry, } x & = 3 \\ \\ \text{Turning point: } & (3, 0) \end{align}
Question 6
(a)
\begin{align} y & = (x - p)(x - q) \phantom{000000} [\text{Minimum curve } \cup] \\ \\ \text{When } & y = 0, x = -1 \text{ or } x = 2 \\ \\ x & = -1 \phantom{00} \text{ or } \phantom{0000} x = 2 \\ x + 1 & = 0 \phantom{000000.} x - 2 = 0 \\ \\ & (x + 1)(x - 2) = 0 \\ \\ \therefore y & = (x + 1)(x - 2) \end{align}
(b)
\begin{align} \text{Line of sym} & \text{metry, } x = -2 \\ \\ \text{Let } a \text{ represent} & \text{ the other } x \text{-intercept} \\ \\ -2 & = {-6 + a \over 2} \\ {-4 \over 2} & = {-6 + a \over 2} \\ -4 & = -6 + a \\ -4 + 6 & = a \\ 2 & = a \\ \\ \\ y & = -(x - p)(x - q) \phantom{000000} [\text{Maximum curve } \cap] \\ \\ \text{When } & y = 0, x = -6 \text{ or } x = 2 \\ \\ x & = -6 \phantom{00} \text{ or } \phantom{0000} x = 2 \\ x + 6 & = 0 \phantom{000000.} x - 2 = 0 \\ \\ & (x + 6)(x - 2) = 0 \\ \\ \therefore y & = -(x + 6)(x - 2) \end{align}
Question 7
(a)
\begin{align} y & = - (x - h)^2 + k \phantom{000000} [\text{Maximum curve}] \\ \\ \text{Turning } & \text{point: } (5, -2) \\ \\ y & = -(x - 5)^2 - 2 \end{align}
(b)
\begin{align} y & = (x - h)^2 + k \phantom{000000} [\text{Minimum curve}] \\ \\ \text{Let } a \text{ rep} & \text{resent the } y \text{-coordinate of the turning point} \\ \\ \text{Turning } & \text{point: } (-2, a) \\ \\ y & = (x + 2)^2 + a \\ \\ \text{Using } & (0, 5), \phantom{0000000000000} [y \text{-intercept}] \\ 5 & = (0 + 2)^2 + a \\ 5 & = 4 + a \\ 5 - 4 & = a \\ 1 & = a \\ \\ \therefore y & = (x + 2)^2 + 1 \end{align}
Question 8
(a)
\begin{align} y & = x^2 + 2x + 3 \phantom{0000000000000000000} [\text{Minimum curve } \cup] \\ y & = x^2 + 2x + \left(2 \over 2\right)^2 - \left(2 \over 2\right)^2 + 3 \\ y & = x^2 + 2x + 1^2 - 1^2 + 3 \\ y & = (x + 1)^2 - 1 + 3 \\ y & = (x + 1)^2 + 2 \\ \\ \text{Turning } & \text{point: } (-1, 2) \\ \\ \text{Line of } & \text{symmetry, } x = -1 \\ \\ \text{Let } & x = 0, \\ y & = (0 + 1)^2 + 2 \\ y & = 3 \phantom{00000000000000} [y \text{-intercept}] \end{align}
(b)
\begin{align} y & = - x^2 + 8x - 5 \phantom{0000000000000000000} [\text{Maximum curve } \cap] \\ y & = -(x^2 - 8x) - 5 \\ y & = - \left[ x^2 - 8x + \left(8 \over 2\right)^2 - \left(8 \over 2\right)^2 \right] - 5 \\ y & = - ( x^2 - 8x + 4^2 - 4^2 ) - 5 \\ y & = - [ (x - 4)^2 - 16 ] - 5 \\ y & = - (x - 4)^2 + 16 - 5 \\ y & = - (x - 4)^2 + 11 \\ \\ \text{Turning } & \text{point: } (4, 11) \\ \\ \text{Line of } & \text{symmetry, } x = 4 \\ \\ \text{Let } & x = 0, \\ y & = -(0 - 4)^2 + 11 \\ y & = -5 \phantom{00000000000000} [y \text{-intercept}] \end{align}
(c)
\begin{align} y & = 5x - x^2 \\ y & = -x^2 + 5x \phantom{0000000000000000000} [\text{Maximum curve } \cap] \\ y & = -(x^2 - 5x) \\ y & = -\left[ x^2 - 5x + \left(5 \over 2\right)^2 - \left(5 \over 2\right)^2 \right] \\ y & = - (x^2 - 5x + 2.5^2 - 2.5^2) \\ y & = - [ (x - 2.5)^2 - 6.25 ] \\ y & = - (x - 2.5)^2 + 6.25 \\ \\ \text{Turning } & \text{point: } (2.5, 6.25) \\ \\ \text{Line of } & \text{symmetry, } x = 2.5 \\ \\ \text{Let } & x = 0, \\ y & = - (0 - 2.5)^2 + 6.25 \\ y & = 0 \phantom{0000000000000000000000000} [y \text{-intercept}] \end{align}
(d)
\begin{align} y & = x^2 - 7x + 6 \phantom{0000000000000000000} [\text{Minimum curve } \cup] \\ y & = x^2 - 7x + \left(7 \over 2\right)^2 - \left(7 \over 2\right)^2 + 6 \\ y & = x^2 - 7x + 3.5^2 - 3.5^2 + 6 \\ y & = (x - 3.5)^2 - 12.25 + 6 \\ y & = (x - 3.5)^2 - 6.25 \\ \\ \text{Turning } & \text{point: } (3.5, -6.25) \\ \\ \text{Line of } & \text{symmetry, } x = 3.5 \\ \\ \text{Let } & x = 0, \\ y & = (0 - 3.5)^2 - 6.25 \\ y & = 6 \phantom{0000000000000000000000000000} [y \text{-intercept}] \end{align}
(e)
\begin{align} y & = x^2 + 4x + 4 \phantom{0000000000000000000} [\text{Minimum curve } \cup] \\ y & = x^2 + 4x + \left(4 \over 2\right)^2 - \left(4 \over 2\right)^2 + 4 \\ y & = x^2 + 4x + 2^2 - 2^2 + 4 \\ y & = (x + 2)^2 - 4 + 4 \\ y & = (x + 2)^2 + 0 \\ \\ \text{Turning } & \text{point: } (-2, 0) \\ \\ \text{Line of } & \text{symmetry, } x = -2 \\ \\ \text{Let } & x = 0, \\ y & = (0 + 2)^2 + 0 \\ y & = 4 \phantom{0000000000000000000000000000} [y \text{-intercept}] \end{align}
(f)
\begin{align} y & = -x^2 - 10x - 25 \phantom{0000000000000000000} [\text{Maximum curve } \cap] \\ y & = - (x^2 + 10x) - 25 \\ y & = - \left[ x^2 + 10x + \left(10 \over 2\right)^2 - \left(10 \over 2\right)^2 \right] - 25 \\ y & = - ( x^2 + 10x + 5^2 - 5^2 ) - 25 \\ y & = - [ (x + 5)^2 - 25 ] - 25 \\ y & = - (x + 5)^2 + 25 - 25 \\ y & = - (x + 5)^2 \\ \\ \text{Turning } & \text{point: } (-5, 0) \\ \\ \text{Line of } & \text{symmetry, } x = -5 \\ \\ \text{Let } & x = 0, \\ y & = - (0 + 5)^2 \\ y & = -25 \phantom{0000000000000000000000000000} [y \text{-intercept}] \end{align}
Question 9
(a)
\begin{align} y & = x^2 - 2x - 8 \\ \\ \text{Let } & x = 0, \\ y & = (0)^2 - 2(0) - 8 \\ y & = -8 \\ \\ \therefore & \phantom{.} C(0, -8) \end{align}
(b)
\begin{align} y & = x^2 - 2x - 8 \\ \\ \text{Let } & y = 0, \\ 0 & = x^2 - 2x - 8 \\ 0 & = (x + 2)(x - 4) \end{align} \begin{align} x + 2 & = 0 & & \text{ or } & x - 4 & =0 \\ x & = - 2 &&& x & = 4 \\ \\ \therefore & \phantom{.} A(-2, 0) &&& \therefore & \phantom{.} B(4, 0) \end{align}
(c)
\begin{align} \text{Line of symmetry, } x & = {-2 + 4 \over 2} \\ x & = 1 \\ \\ \text{Substitute } & x = 1 \text{ into eqn of curve,} \\ y & = (1)^2 - 2(1) - 8 \\ y & = -9 \\ \\ \text{Minimum } & \text{point: } (1, -9) \end{align}
Question 10
(a)
\begin{align} y & = x^2 + 2x + k \\ \\ \text{Using } & P(-3, 0), \\ 0 & = (-3)^2 + 2(-3) + k \\ 0 & = 9 - 6 + k \\ 0 & = 3 + k \\ -3 & = k \end{align}
(b)
\begin{align} y & = x^2 + 2x + k \\ y & = x^2 + 2x - 3 \\ \\ \text{Let } & x = 0, \\ y & = (0)^2 + 2(0) - 3 \\ y & = -3 \\ \\ \therefore & \phantom{.} R(0, -3) \\ \\ \\ \text{Let } & y = 0, \\ 0 & = x^2 + 2x - 3 \\ 0 & = (x + 3)(x - 1) \end{align} \begin{align} x + 3 & = 0 & & \text{ or } & x - 1 & =0 \\ x & = - 3 &&& x & = 1 \\ \\ \therefore & \phantom{.} P(-3, 0) &&& \therefore & \phantom{.} Q(1, 0) \end{align}
(c)
\begin{align} \text{Line of symmetry, } x & = {-3 + 1 \over 2} \\ x & = -1 \end{align}
Question 11
(a)
\begin{align} y & = -2x^2 + 48x - 160 \phantom{0000000000} [\text{Maximum curve } \cap] \\ y & = -2(x^2 - 24x) - 160 \\ y & = -2 \left[ x^2 - 24x + \left(24 \over 2\right)^2 - \left(24 \over 2\right)^2 \right] - 160 \\ y & = -2 ( x^2 - 24x + 12^2 - 12^2 ) - 160 \\ y & = -2 [ (x - 12)^2 - 144 ] - 160 \\ y & = -2 (x - 12)^2 + 288 - 160 \\ y & = -2 (x - 12)^2 + 128 \\ \\ \text{Turning } & \text{point: } (12, 128) \\ \\ \text{Let } & x = 0, \\ y & = -2(0 - 12)^2 + 128 \\ y & = -160 \phantom{000000000000000000000} [y \text{-intercept}] \end{align}
(b)
\begin{align} \text{From (a), } & y = -160 \\ \\ \text{Sam loses \$ 160 daily } & \text{if he doesn't make any vase} \end{align}
(c)(i)
\begin{align} \text{Turning point} & \text{: } (12, 128) \\ \\ \text{No. of vases }(x) & = 12 \end{align}
(c)(ii)
\begin{align} \text{Turning point} & \text{: } (12, 128) \\ \\ \text{Max. daily profit } (y) & = \$ 128 \end{align}
Question 12
(a)
\begin{align}
\text{Let coordinates} & \text{ of } R \text{ be } (r, 0) \\
\\
\text{Line of symmetry, } x & = {-5 + r \over 2} \\
1 & = {-5 + r \over 2} \\
{2 \over 2} & = {-5 + r \over 2} \\
2 & = -5 + r \\
2 + 5 & = r \\
7 & = r \\
\\
\therefore & \phantom{.} R(7, 0)
\end{align}
\begin{align}
x & = -5
&& \text{ or } &
x & = 7 \\
x + 5 & = 0
&&&
x - 7 & = 0
\end{align}
\begin{align}
0 & = (x + 5)(x - 7) \\
0 & = x^2 - 7x + 5x - 35 \\
0 & = x^2 - 2x - 35 \\
\\
\text{Eqn of curve: } y & = -(x^2 - 2x - 35) \\
y & = - x^2 + 2x + 35 \\
\\
\text{Let } & x = 0, \\
y & = -(0)^2 + 2(0) + 35 \\
y & = 35 \\
\\
\therefore & \phantom{.} Q(0, 35)
\end{align}
(b)
\begin{align} \text{Line of symmetry, } & x = 1 \\ \\ \text{Substitute } x = 1 & \text{ into eqn of curve,} \\ y & = -(1)^2 + 2(1) + 35 \\ y & = 36 \\ \\ \text{Max. point: } & (1, 36) \end{align}
(c)
\begin{align} \text{Substitute } x = -3 & \text{ into eqn of curve,} \\ y & = -(-3)^2 + 2(-3) + 35 \\ y & = 20 \\ \\ \therefore k & = 20 \end{align}
Question 13
(a)
\begin{align} x + 7 \end{align}
(b)
\begin{align} \text{Product} & = x \times (x + 7) \\ y & = x(x + 7) \\ y & = x^2 + 7x \end{align}
(c)
\begin{align} y & = x^2 + 7x \phantom{0000000000} [\text{Minimum curve } \cup] \\ y & = x^2 + 7x + \left(7 \over 2\right)^2 - \left(7 \over 2\right)^2 \\ y & = x^2 + 7x + 3.5^2 - 3.5^2 \\ y & = (x + 3.5)^2 - 12.25 \\ \\ \text{Turning } & \text{point: } (-3.5, -12.25) \\ \\ \text{Let } & x = 0, \\ y & = (0 + 3.5)^2 - 12.25 \\ y & = 0 \phantom{000000000000000000000} [y \text{-intercept}] \end{align}
(d)
\begin{align} \text{Turning point: } & (-3.5, -12.25) \\ \\ \text{Min. value of product } (y) & = -12.25 \\ \\ \text{Smaller number } (x) & = -3.5 \\ \\ \text{Larger number} & = x + 7 \\ & = -3.5 + 7 \\ & = 3.5 \end{align}
Question 14
(a)
(x = 20 is the line of symmetry)
\begin{align} \text{Distance from } O & = 2 \times 20 \\ & = 40 \text{ m} \end{align}
(b)
\begin{align} x & = 0 &&& x & = 40 \\ & &&& x - 40 & = 0 \end{align} \begin{align} 0 & = x(x - 40) \\ \\ \text{Eqn of curve: } & y = a x (x - 40) \\ \\ \text{Using} & \text{ max. point } (20, 5), \\ 5 & = a(20)(20 - 40) \\ 5 & = a(20)(-20) \\ 5 & = -400a \\ {5 \over -400} & = a \\ -{1 \over 80} & = a \\ \\ \text{Eqn of curve: } & y = -{1 \over 80} x (x - 40) \\ \end{align}
Question 15
(a)
\begin{align} y & = 2 + 4x - x^2 \\ \\ \text{Let } & x = 0, \\ y & = 2 + 4(0) - (0)^2 \\ y & = 2 \\ \\ OA & = 2 \text{ m} \end{align}
(b)
\begin{align}
y & = 2 + 4x - x^2 \\
\\
\text{Let } & y = 2, \\
2 & = 2 + 4x - x^2 \\
2 - 2 & = 4x - x^2 \\
0 & = 4x - x^2 \\
0 & = x(4 - x)
\end{align}
\begin{align}
x & = 0
&& \text{ or } &
4 - x & = 0 \\
& &&&
-x & = -4 \\
& &&&
x & = 4
\end{align}
\begin{align}
\therefore C(4, 2) & \text{ and } D(4, 0) \\
\\
OD & = 4 \text{ m}
\end{align}
(c)
\begin{align} \text{Line of symmetry, } x & = {0 + 4 \over 2} \\ x & = 2 \\ \\ \text{Substitute } x = 2 & \text{ into } y = 2 + 4x - x^2, \\ y & = 2 + 4(2) - (2)^2 \\ y & = 6 \\ \\ \therefore & \phantom{.} B(2, 6) \end{align}
(d)
\begin{align} y & = 2 + 4x - x^2 \\ y & = - x^2 + 4x + 2 \phantom{000000} [\text{Maximum curve } \cap ] \\ \\ y & \text{-intercept: } 2 \\ \\ \text{Ends of } & \text{arch: } A(0, 2), C(4, 2) \\ \\ \text{Tur} & \text{ning point: } (2, 6) \end{align}
Question 16
(a)
\begin{align} y & = \pm (x - h)^2 + k \\ \\ \text{Since tur} & \text{ning point is } (3, - 1), \\ y & = \pm (x - 3)^2 - 1 \end{align}
(b)
\begin{align} y & = (x - 3)^2 - 1 \phantom{000000} [\text{Minimum curve } \cup] \\ \\ \text{Turning} & \text{ point: } (3, -1) \\ \\ \text{Let } & x = 0, \\ y & = (0 - 3)^2 - 1 \\ y & = 8 \phantom{000000000000000} [y \text{-intercept}] \end{align}
\begin{align} y & = -(x - 3)^2 - 1 \phantom{000000} [\text{Maximum curve } \cap] \\ \\ \text{Turning} & \text{ point: } (3, -1) \\ \\ \text{Let } & x = 0, \\ y & = -(0 - 3)^2 - 1 \\ y & = -10 \phantom{000000000000000} [y \text{-intercept}] \end{align}
Question 17
(a)
\begin{align} y & = (x - 2)^2 + (c - 4) \\ \\ \text{Turning} & \text{ point: } (2, c - 4) \end{align} \begin{align} h & = 2 &&& c - 4 & = -5 \\ & &&& c & = -5 + 4 \\ & &&& c & = -1 \end{align}
(b)
\begin{align} y & = x^2 - 4x + c \\ y & = x^2 - 4x - 1 \\ y & = x^2 - 4x + \left(4 \over 2\right)^2 - \left(4 \over 2\right)^2 - 1 \\ y & = (x - 2)^2 - 4 - 1 \\ y & = (x - 2)^2 - 5 \\ \\ \text{Minimum} & \text{ point: } (2, -5) \\ \\ \text{Let } & x = 0, \\ y & = (0 - 2)^2 - 5 \\ y & = -1 \phantom{000000000000} [y \text{-intercept}] \end{align}
Question 18
(a)
\begin{align} \text{If Becky is correct, } y & = (x + 2)(x - 3) \\ & = x^2 - x - 6 \\ \\ \text{Line of symmetry, } x & = {-2 + 3 \over 2} = 0.5 \neq 1 \phantom{00} [\text{Contradicts Alice}] \\ \\ y \text{-intercept} & = -6 \neq 4 \phantom{00} [\text{Contradicts Casey}] \\ \\ \therefore \text{Becky is wrong, } & \text{Alice and Casey are correct} \end{align}
(b)
\begin{align} y & = x^2 - bx + c \\ \\ \text{Using } & (0, 4), \\ 4 & = (0)^2 - b(0) + c \\ 4 & = 0 - 0 + c \\ 4 & = c \\ \\ y & = x^2 - bx + 4 \\ y & = x^2 - bx + \left(b \over 2\right)^2 - \left(b \over 2\right)^2 + 4 \\ y & = \left(x - {b \over 2}\right)^2 - {b^2 \over 4} + 4 \\ \\ \text{Minimum} & \text{ point: } \left({b \over 2}, -{b^2 \over 4} + 4\right) \\ \\ \therefore {b \over 2} & = 1 \\ {b \over 2} & = {2 \over 2} \\ b & = 2 \\ \\ \\ \therefore b & = 2, c = 4 \end{align}
Review Exercise 1
Question 1
(a)
\begin{align} 2x^2 + x - 15 & = 0 \\ (2x - 5)(x + 3) & = 0 \end{align} \begin{align} 2x - 5 & = 0 && \text{ or } & x + 3 & = 0 \\ 2x & = 5 &&& x & = -3 \\ x & = {5 \over 2} \end{align}
(b)
\begin{align} 3x^2 - 20x - 7 & = 0 \\ (3x + 1)(x - 7) & = 0 \end{align} \begin{align} 3x + 1 & = 0 && \text{ or } & x - 7 & = 0 \\ 3x & = -1 &&& x & = 7 \\ x & = -{1 \over 3} \end{align}
(c)
\begin{align} 9x^2 - 25 & = 0 \\ (3x)^2 - (5)^2 & = 0 \\ (3x + 5)(3x - 5) & = 0 \phantom{000000} [a^2 - b^2 = (a + b)(a - b)] \end{align} \begin{align} 3x + 5 & = 0 && \text{ or } & 3x - 5 & = 0 \\ 3x & = -5 &&& 3x & = 5 \\ x & = -{5 \over 3} &&& x & = {5 \over 3} \end{align}
(d)
\begin{align} (x + 3)(x - 2) & = 5x(x - 2) \\ x^2 - 2x + 3x - 6 & = 5x^2 - 10x \\ x^2 + x - 6 & = 5x^2 - 10x \\ 0 & = 5x^2 - x^2 - 10x - x + 6 \\ 0 & = 4x^2 - 11x + 6 \\ 0 & = (4x - 3)(x - 2) \end{align} \begin{align} 4x - 3 & = 0 && \text{ or } & x - 2 & = 0 \\ 4x & = 3 &&& x & = 2 \\ x & = {3 \over 4} \end{align}
Question 2
(a)
\begin{align} x^2 - 6x + 4 & = 0 \\ x^2 - 6x + \left(6 \over 2\right)^2 - \left(6 \over 2\right)^2 + 4 & = 0 \\ x^2 - 6x + 3^2 - 3^2 + 4 & = 0 \\ (x - 3)^2 - 9 + 4 & = 0 \\ (x - 3)^2 - 5 & = 0 \\ (x - 3)^2 & = 5 \\ x - 3 & = \pm \sqrt{5} \end{align} \begin{align} x - 3 & = \sqrt{5} && \text{ or } & x - 3 & = - \sqrt{5} \\ x & = \sqrt{5} + 3 &&& x & = - \sqrt{5} + 3 \\ x & \approx 5.24 &&& x & \approx 0.764 \end{align}
(b)
\begin{align} x^2 + 4x - 3 & = 0 \\ x^2 + 4x + \left(4 \over 2\right)^2 - \left(4 \over 2\right)^2 - 3 & =0 \\ x^2 + 4x + 2^2 - 2^2 - 3 & = 0 \\ (x + 2)^2 - 4 - 3 & = 0 \\ (x + 2)^2 - 7 & = 0 \\ (x + 2)^2 & = 7 \\ x + 2 & = \pm \sqrt{7} \end{align} \begin{align} x + 2 & = \sqrt{7} && \text{ or } & x + 2 & = - \sqrt{7} \\ x & = \sqrt{7} - 2 &&& x & = - \sqrt{7} - 2 \\ x & \approx 0.646 &&& x & \approx -4.65 \end{align}
(c)
\begin{align} x^2 + 15x + 8 & = 0 \\ x^2 + 15x + \left(15 \over 2\right)^2 - \left(15 \over 2\right)^2 + 8 & = 0 \\ x^2 + 15x + 7.5^2 - 7.5^2 + 8 & = 0 \\ (x + 7.5)^2 - 56.25 + 8 & = 0 \\ (x + 7.5)^2 - 48.25 & = 0 \\ (x + 7.5)^2 & = 48.25 \\ x + 7.5 & = \pm \sqrt{48.25} \end{align} \begin{align} x + 7.5 & = \sqrt{48.25} && \text{ or } & x + 7.5 & = - \sqrt{48.25} \\ x & = \sqrt{48.25} - 7.5 &&& x & = - \sqrt{48.25} - 7.5 \\ x & \approx -0.554 &&& x & \approx -14.4 \end{align}
(d)
\begin{align} x^2 - 7x - 2 & = 0 \\ x^2 - 7x + \left(7 \over 2\right)^2 - \left(7 \over 2\right)^2 - 2 & = 0 \\ x^2 - 7x + 3.5^2 - 3.5^2 - 2 & = 0 \\ (x - 3.5)^2 - 12.25 - 2 & =0 \\ (x - 3.5)^2 - 14.25 & = 0 \\ (x - 3.5)^2 & = 14.25 \\ x - 3.5 & = \pm \sqrt{14.25} \end{align} \begin{align} x - 3.5 & = \sqrt{14.25} && \text{ or } & x - 3.5 & = - \sqrt{14.25} \\ x & = \sqrt{14.25} + 3.5 &&& x & = - \sqrt{14.25} + 3.5 \\ x & \approx 7.27 &&& x & \approx -0.275 \end{align}
Question 3
(a)
\begin{align} x^2 & + 13x - 88 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-13 \pm \sqrt{(13)^2 - 4(1)(-88)} \over 2(1)} \\ & = {-13 \pm \sqrt{521} \over 2} \\ & = 4.9127 \text{ or } -17.912 \\ & \approx 4.91 \text{ or } -17.9 \end{align}
(b)
\begin{align} 3x^2 & + 12x + 1 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-12 \pm \sqrt{(12)^2 - 4(3)(1)} \over 2(3)} \\ & = {-12 \pm \sqrt{132} \over 6} \\ & = -0.085 \phantom{.} 145 \text{ or } -3.9148 \\ & \approx -0.0851 \text{ or } -3.91 \end{align}
(c)
\begin{align} -4x^2 & + 5x - 9 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-5 \pm \sqrt{(5)^2 - 4(-4)(-9)} \over 2(-4)} \\ & = {-5 \pm \sqrt{-119} \over -8} \\ \\ \\ & \sqrt{-119} \text{ is not a real number} \\ \\ \therefore & \text{ Equation has no real roots} \end{align}
(d)
\begin{align} -5x^2 & + 8x + 2 = 0 \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-8 \pm \sqrt{(8)^2 - 4(-5)(2)} \over 2(-5)} \\ & = {-8 \pm \sqrt{104} \over -10} \\ & = -0.21980 \text{ or } 1.8198 \\ & \approx -0.220 \text{ or } 1.82 \end{align}
Question 4
(a)
\begin{align} {x + 1 \over 2} & = {8 \over x + 1} \\ {(x + 1)(x + 1) \over 2(x + 1)} & = {8(2) \over 2(x + 1)} \\ \\ (x + 1)(x + 1) & = 8(2) \\ x^2 + x + x + 1 & = 16 \\ x^2 + 2x + 1 & = 16 \\ x^2 + 2x + 1 - 16 & = 0 \\ x^2 + 2x - 15 & = 0 \\ (x + 5)(x - 3) & =0 \end{align} \begin{align} x + 5 & = 0 && \text{ or } & x - 3 & =0 \\ x & = -5 &&& x & = 3 \end{align}
(b)
\begin{align} {10(x - 2) \over 1} + {10 \over x - 2} & = {29 \over 1} \\ {10(x - 2)(x - 2) \over x - 2} + {10 \over x - 2} & = {29(x - 2) \over x - 2} \\ {10(x - 2)(x - 2) + 10 \over x - 2} & = {29(x - 2) \over x - 2} \\ \\ 10(x - 2)(x - 2) + 10 & = 29(x - 2) \\ 10(x^2 - 2x - 2x + 4) + 10 & = 29x - 58 \\ 10(x^2 - 4x + 4) + 10 & = 29x - 58 \\ 10x^2 - 40x + 40 + 10 & = 29x - 58 \\ 10x^2 - 40x + 50 & = 29x - 58 \\ 10x^2 - 40x - 29x + 50 + 58 & = 0 \\ 10x^2 - 69x + 108 & = 0 \\ (5x - 12)(2x - 9) & = 0 \end{align} \begin{align} 5x - 12 & = 0 && \text{ or } & 2x - 9 & = 0 \\ 5x & = 12 &&& 2x & = 9 \\ x & = {12 \over 5} &&& x & = {9 \over 2} \end{align}
(c)
\begin{align} {x \over x + 1} + {x + 1 \over x} & = {3 \over 1} \\ {x^2 \over x(x + 1)} + {(x + 1)(x + 1) \over x(x + 1)} & = {3x(x + 1) \over x(x + 1)} \\ {x^2 + (x + 1)(x + 1) \over x(x + 1)} & = {3x(x + 1) \over x(x + 1)} \\ \\ x^2 + (x + 1)(x + 1) & = 3x(x + 1) \\ x^2 + x^2 + x + x + 1 & = 3x^2 + 3x \\ 2x^2 + 2x + 1 & = 3x^2 + 3x \\ 0 & = 3x^2 - 2x^2 + 3x - 2x - 1 \\ 0 & = x^2 + x - 1 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-1 \pm \sqrt{(1)^2 - 4(1)(-1)} \over 2(1)} \\ & = {-1 \pm \sqrt{5} \over 2} \\ & = 0.61803 \text{ or } -1.618 \\ & \approx 0.618 \text{ or } -1.62 \end{align}
(d)
\begin{align} {3 \over 7x - 2} - {2 \over 7x + 3} & = {3 \over 4} \\ {3(4)(7x + 3) \over 4(7x - 2)(7x + 3)} - {2(4)(7x - 2) \over 4(7x - 2)(7x + 3)} & = {3(7x - 2)(7x + 3) \over 4(7x - 2)(7x + 3)} \\ {12(7x + 3) - 8(7x - 2) \over 4(7x - 2)(7x + 3)} & = {3(7x - 2)(7x + 3) \over 4(7x - 2)(7x + 3)} \\ \\ 12(7x + 3) - 8(7x - 2) & = 3(7x - 2)(7x + 3) \\ 84x + 36 - 56x + 16 & = 3(49x^2 + 21x - 14x - 6) \\ 28x + 52 & = 3(49x^2 + 7x - 6) \\ 28x + 52 & = 147x^2 + 21x - 18 \\ 0 & = 147x^2 + 21x - 28x - 18 - 52 \\ 0 & = 147x^2 - 7x - 70 \\ 0 & = 21x^2 - x - 10 \\ 0 & = (3x + 2)(7x - 5) \end{align} \begin{align} 3x + 2 & = 0 && \text{ or } & 7x - 5 & = 0 \\ 3x & = -2 &&& 7x & = 5 \\ x & = -{2 \over 3} &&& x & = {5 \over 7} \end{align}
Question 5
(a)
\begin{align} 3x^2 & + 7x + 1 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-7 \pm \sqrt{(7)^2 - 4(3)(1)} \over 2(3)} \\ & = {-7 \pm \sqrt{37} \over 6} \\ & = -0.15287 \text{ or } -2.1805 \\ & \approx -0.153 \text{ or } -2.18 \end{align}
(b)
\begin{align} (2x - 1)^2 & = (x + 1)^2 \\ 2x - 1 & = \pm \sqrt{(x + 1)^2} \\ 2x - 1 & = \pm (x + 1) \end{align} \begin{align} 2x - 1 & = x + 1 && \text{ or } & 2x - 1 & = -(x + 1) \\ 2x - x & = 1 + 1 &&& 2x - 1 & = -x - 1 \\ x & = 2 &&& 2x + x & = -1 + 1 \\ & &&& 3x & = 0 \\ & &&& x & = 0 \end{align}
(c)
\begin{align} (x + 3)(2x - 9) & = x(x + 2) \\ 2x^2 - 9x + 6x - 27 & = x^2 + 2x \\ 2x^2 - 3x - 27 & = x^2 + 2x \\ 2x^2 - x^2 - 3x - 2x - 27 & = 0 \\ x^2 - 5x - 27 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-5) \pm \sqrt{(-5)^2 - 4(1)(-27)} \over 2(1)} \\ & = {5 \pm \sqrt{133} \over 2} \\ & = 8.2663 \text{ or } -3.2663 \\ & \approx 8.27 \text{ or } -3.27 \end{align}
(d)
\begin{align} {1 \over 3x} + {2 \over 3x - 1} + {3 \over 3x + 1} & = {0 \over 1} \\ {(3x - 1)(3x + 1) \over 3x(3x - 1)(3x + 1)} + {2(3x)(3x + 1) \over 3x(3x - 1)(3x + 1)} + {3(3x)(3x - 1) \over 3x(3x - 1)(3x + 1)} & = {0 \over 3x(3x - 1)(3x + 1)} \\ {(3x - 1)(3x + 1) + 6x(3x + 1) + 9x(3x - 1) \over 3x(3x - 1)(3x + 1)} & = {0 \over 3x(3x - 1)(3x + 1)} \\ (3x - 1)(3x + 1) + 6x(3x + 1) + 9x(3x - 1) & = 0 \\ 9x^2 + 3x - 3x - 1 + 18x^2 + 6x + 27x^2 - 9x & = 0 \\ 54x^2 - 3x - 1 & = 0 \\ (9x + 1)(6x - 1) & = 0 \end{align} \begin{align} 9x + 1 & = 0 && \text{ or } & 6x - 1 & = 0 \\ 9x & = -1 &&& 6x & = 1 \\ x & = -{1 \over 9} &&& x & = {1 \over 6} \end{align}
Question 6
(a)
\begin{align} x^2 + 8x - 5 & = x^2 + 8x + \left(8 \over 2\right)^2 - \left(8 \over 2\right)^2 - 5 \phantom{000000} [\text{Complete the square}] \\ & = x^2 + 8x + 4^2 - 4^2 - 5 \\ & = (x + 4)^2 - 16 - 5 \\ & = (x + 4)^2 - 21 \end{align}
(b)
\begin{align} x^2 + 8x - 5 & = 0 \\ (x + 4)^2 - 21 & = 0 \\ (x + 4)^2 & = 21 \\ x + 4 & = \pm \sqrt{21} \end{align} \begin{align} x + 4 & = \sqrt{21} && \text{ or } & x + 4 & = - \sqrt{21} \\ x & = \sqrt{21} - 4 &&& x & = - \sqrt{21} - 4 \\ x & \approx 0.583 &&& x & \approx -8.58 \end{align}
(c)
\begin{align} y & = x^2 + 8x - 5 \phantom{000000} [\text{Minimum curve } \cup] \\ y & = (x + 4)^2 - 21 \\ \\ \text{Turning} & \text{ point: } (-4, -21) \\ \\ \text{Line of symmetry, } & x = -4 \\ \\ \text{Let } & x = 0, \\ y & = (0 + 4)^2 - 21 \\ y & = -5 \phantom{0000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = (x + 4)^2 - 21 \\ \\ \text{From (b), } & x \approx 0.583, -8.58 \phantom{0000} [x \text{-intercepts}] \end{align}
Question 7
(a)
\begin{align} -x^2 + 12x - 35 & = -(x^2 - 12x + 35) \\ & = - (x - 5)(x - 7) \end{align}
(b)
\begin{align} y & = -x^2 + 12x - 35 \phantom{000000} [\text{Maximum curve } \cap] \\ y & = - (x - 5)(x - 7) \\ \\ \text{Let } & x = 0, \\ y & = -(0 - 5)(0 - 7) \\ y & = - 35 \phantom{0000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = - (x - 5)(x - 7) \\ \\ x - 5 & = 0 \phantom{0} \text{ or } \phantom{0} x - 7 = 0 \\ x & = 5 \phantom{00000000} x = 7 \phantom{000000} [x \text{-intercepts}] \\ \\ \text{Line of symmetry, } x & = {5 + 7 \over 2} \\ x & = 6 \\ \\ \text{Let } & x = 6, \\ y & = -(6 - 5)(6 - 7) \\ y & = 1 \\ \\ \text{Turning} & \text{ point: } (6, 1) \end{align}
Question 8
(a)
| x | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 |
|---|---|---|---|---|---|---|---|---|
| y | 4 | -2 | -6 | -8 | -8 | -6 | -2 | 4 |
(b)
\begin{align} \underbrace{x^2 + 3x - 6}_\text{Curve} & = 0 \\ \\ \text{From graph, } & x = -4.4, 1.4 \phantom{000000} [\text{Look for } x \text{-intercepts}] \end{align}
(c)(i)
\begin{align} x^2 + 3x - 3 & = 0 \\ x^2 + 3x - 3 - 3 & = 0 - 3 \\ \underbrace{x^2 + 3x - 6}_\text{Curve} & = -3 \\ \\ \text{Draw } & y = -3 \\ \\ \text{From graph, } & x = -3.8, 0.8 \end{align}
(c)(ii)
\begin{align} x^2 + x - 5 & = 0 \\ x^2 + x - 5 + 2x & = 0 + 2x \\ x^2 + 3x - 5 & = 2x \\ x^2 + 3x - 5 - 1 & = 2x - 1 \\ \underbrace{x^2 + 3x - 6}_\text{Curve} & = 2x - 1 \\ \\ \text{Draw } & y = 2x - 1 \end{align}
| x | -1 | 0 | 1 |
|---|---|---|---|
| y | -3 | -1 | 1 |
$$ \text{From graph, } x = -2.8, 1.8 $$
Question 9
(a)
| x | -3 | -2 | -1 | 0 | 1 | 2 |
|---|---|---|---|---|---|---|
| y | -7 | 1 | 5 | 5 | 1 | -7 |
(b)
\begin{align} \underbrace{5 - 2x - 2x^2}_\text{Curve} & = 0 \\ \\ \text{From graph, } & x = -2.15, 1.15 \phantom{000000} [\text{Look for } x \text{-intercepts}] \end{align}
(c)(i)
\begin{align} 1 - 2x - 2x^2 & = 0 \\ 1 - 2x - 2x^2 + 4 & = 0 + 4 \\ \underbrace{5 - 2x - 2x^2}_\text{Curve} & = 4 \\ \\ \text{Draw } & y = 4 \\ \\ \text{From graph, } & x = -1.35, 0.35 \end{align}
(c)(ii)
\begin{align} 8 - x - 2x^2 & = 0 \\ 8 - x - 2x^2 - 3 & = 0 - 3 \\ 5 - x - 2x^2 & = - 3 \\ 5 - x - 2x^2 - x & = - 3 - x \\ \underbrace{5 - 2x - 2x^2}_\text{Curve} & = - x - 3 \\ \\ \text{Draw } & y = - x - 3 \end{align}
| x | -1 | 0 | 1 |
|---|---|---|---|
| y | -2 | -3 | -4 |
$$ \text{From graph, } x = -2.25, 1.8 $$
Question 10
(a)
\begin{align}
y & = x^2 + 4x \\
\\
\text{Let } & y = 0, \\
0 & = x^2 + 4x \\
0 & = x(x + 4)
\end{align}
\begin{align}
x & = 0
&& \text{ or } &
x + 4 & = 0 \\
&
&&&
x & = -4
\end{align}
$$ \therefore A(-4, 0) $$
(b)
\begin{align} x & = {-4 + 0 \over 2} \\ x & = -2 \end{align}
(c)
\begin{align} \text{Substitute } & x = - 2 \text{ into } y = x^2 + 4x, \\ y & = (-2)^2 + 4(-2) \\ y & = -4 \\ \\ \text{Minimum} & \text{ point: } (-2, -4) \end{align}
Question 11
\begin{align}
\text{Area of lawn} & = 2x \times x \\
& = 2x^2 \text{ m}^2 \\
\\
\text{Area of 1 path} & = 1.5 \times x \\
& = 1.5x \text{ m}^2 \\
\\
\text{Total area} & = 2x^2 + 2(1.5)x \\
189 & = 2x^2 + 3x \\
0 & = 2x^2 + 3x - 189 \\
0 & = (x - 9)(2x + 21)
\end{align}
\begin{align}
x - 9 & = 0
&& \text{ or } &
2x + 21 & = 0 \\
x & = 9
&&&
2x & = -21 \\
& &&&
x & = -{21 \over 2} \text{ (Reject, since } x > 0)
\end{align}
\begin{align}
\text{Breadth of lawn} & = x \\
& = 9 \text{ m} \\
\\
\text{Length of lawn} & = 2x \\
& = 2(9) \\
& = 18 \text{ m}
\end{align}
Question 12
\begin{align} \text{Area of bed} & = 30 \times 20 \\ & = 600 \text{ m}^2 \\ \\ \text{Area of path} & = {1 \over 3} \times 600 \\ & = 200 \text{ m}^2 \\ \\ \text{Let } x \text{ represent } & \text{the width of the path} \\ \\ \text{Length of bed and path} & = (30 + 2x) \text{ m} \\ \\ \text{Breadth of bed and path} & = (20 + 2x) \text{ m} \\ \\ \text{Area of bed and path} & = (30+2x) (20 + 2x) \\ 600 + 200 & = 600 + 60x + 40x + 4x^2 \\ 800 & = 4x^2 + 100x + 600 \\ 0 & = 4x^2 + 100x + 600 - 800 \\ 0 & = 4x^2 + 100x - 200 \\ 0 & = x^2 + 25x - 50 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-25 \pm \sqrt{(25)^2 - 4(1)(-50)} \over 2(1)} \\ & = {-25 \pm \sqrt{825} \over 2} \\ & = 1.8614 \text{ or } -26.861 \text{ (Reject, since } x > 0) \\ \\ \text{Width of path} & = x \\ & = 1.8614 \\ & \approx 1.86 \text{ m} \end{align}
Question 13
\begin{align}
\text{Let } x \text{ denote the ini} & \text{tial speed (in km/h) of the train} \\
\\
\text{Time} & = {\text{Distance} \over \text{Speed}} \\
\\
\text{Time taken (initial)} & = {180 \over x} \text{ h} \\
\\
\text{Time taken (after)} & = {180 \over x + 10} \text{ h} \\
\\
15 \text{ mins} & = {15 \over 60} \text{ h} \\
& = {1 \over 4} \text{ h} \\
\\
{180 \over x} - {180 \over x + 10} & = {1 \over 4} \\
{180(4)(x + 10) \over 4x(x + 10)} - {180(4)(x) \over 4x(x + 10)} & = {x(x + 10) \over 4x(x + 10)} \\
{720(x + 10) \over 4x(x + 10)} - {720x \over 4x(x + 10)} & = {x(x + 10) \over 4x(x + 10)} \\
\\
720(x + 10) - 720x & = x(x + 10) \\
720x + 7200 - 720x & = x^2 + 10x \\
7200 & = x^2 + 10x \\
0 & = x^2 + 10x - 7200 \\
0 & = (x - 80)(x + 90)
\end{align}
\begin{align}
x - 80 & = 0
&& \text{ or } &
x + 90 & = 0 \\
x & = 80
&&&
x & = -90 \text{ (Reject, since } x > 0)
\end{align}
$$ \text{Average speed} = x = 80 \text{ km/h} $$
Question 14
(a)(i)
\begin{align} \text{Time} & = {\text{Distance} \over \text{Speed}} \\ \\ \text{Time taken (Abel)} & = {42 \over x} \text{ h} \end{align}
(a)(ii)
\begin{align} \text{Time taken (Sam)} & = {42 \over x + 3} \text{ h} \end{align}
(b)
\begin{align} 28 \text{ mins} & = {28 \over 60} \text{ h} \\ & = {7 \over 15} \text{ h} \\ \\ {42 \over x} - {42 \over x + 3} & = {7 \over 15} \\ {42(15)(x + 3) \over 15x(x + 3)} - {42(15)(x) \over 15x(x + 3)} & = {7x(x + 3) \over 15x(x + 3)} \\ {630(x + 3) - 630x \over 15x(x + 3)} & = {7x(x + 3) \over 15x(x + 3)} \\ \\ 630(x + 3) - 630x & = 7x(x + 3) \\ 630x + 1890 - 630x & = 7x^2 + 21x \\ 1890 & = 7x^2 + 21x \\ 0 & = 7x^2 + 21x - 1890 \\ 0 & = x^2 + 3x - 270 \phantom{00} \text{ (Shown)} \end{align}
(c)
\begin{align} 0 & = x^2 + 3x - 270 \\ 0 & = (x - 15)(x + 18) \end{align} \begin{align} x - 15 & = 0 && \text{ or } & x + 18 & = 0 \\ x & = 15 &&& x & = -18 \end{align}
(d) Use x = 15, since x (Abel's speed) cannot be a negative value
\begin{align} \text{From (a)(i), time taken (Abel)} & = {42 \over x} \text{ h} \\ & = {42 \over 15} \text{ h} \\ & = 2.8 \text{ h} \\ & = 2 \text{ h } (0.8 \times 60) \text{ mins} \\ & = 2 \text{ h } 48 \text{ mins} \end{align}
Question 15
(a)(i)
\begin{align} x \text{ mins} & \rightarrow 60 \text{ doughnuts} \\ 1 \text{ min} & \rightarrow {60 \over x} \text{ doughnuts} \end{align}
(a)(ii)
\begin{align} (2x - 1) \text{ mins} & \rightarrow 135 \text{ doughnuts} \\ 1 \text{ min} & \rightarrow {135 \over 2x - 1} \text{ doughnuts} \end{align}
(b)
\begin{align} {60 \over x} + {135 \over 2x - 1} & = 27 \\ {60 \over x} + {135 \over 2x - 1} & = {27 \over 1} \\ {60(2x - 1) \over x(2x - 1)} + {135x \over x(2x - 1)} & = {27x(2x - 1) \over x(2x - 1)} \\ {60(2x - 1) + 135x \over x(2x - 1)} & = {27x(2x - 1) \over x(2x - 1)} \\ \\ 60(2x - 1) + 135x & = 27x(2x - 1) \\ 120x - 60 + 135x & = 54x^2 - 27x \\ 255x - 60 & = 54x^2 - 27x \\ 0 & = 54x^2 - 27x - 255x + 60 \\ 0 & = 54x^2 - 282x + 60 \\ 0 & = 9x^2 - 47x + 10 \phantom{00} \text{ (Shown)} \end{align}
(c)
\begin{align} 0 & = 9x^2 - 47x + 10 \\ 0 & = (x - 5)(9x - 2) \end{align} \begin{align} x - 5 & = 0 && \text{ or } & 9x - 2 & = 0 \\ x & = 5 &&& 9x & = 2 \\ & &&& x & = {2 \over 9} \end{align}
(d) Use x = 5 instead of x = 2/9 - if x = 2/9, Machine B will take -5/9 minutes to make 135 doughnuts
\begin{align} \text{From (a)(i), } 1 \text{ min} \rightarrow {60 \over x} & = {60 \over 5} = 12 \text{ doughnuts} \\ \\ \text{Time taken to make 90 doughnuts} & = {90 \over 12} \\ & = 7.5 \text{ mins} \end{align}
Question 16
(a)
$$ {1200 \over x} \text{ mins} $$
(b)
$$ {1200 \over x + 10} \text{ mins} $$
(c)
\begin{align} {1200 \over x} - {1200 \over x + 10} & = 4 \\ {1200 \over x} - {1200 \over x + 10} & = {4 \over 1} \\ {1200(x + 10) \over x(x + 10)} - {1200x \over x(x + 10)} & = {4x(x + 10) \over x(x + 10)} \\ {1200(x + 10) - 1200x \over x(x + 10)} & = {4x(x + 10) \over x(x + 10)} \\ \\ 1200(x + 10) - 1200x & = 4x(x + 10) \\ 1200x + 12 \phantom{.} 000 - 1200x & = 4x^2 + 40x \\ 12 \phantom{.} 000 & = 4x^2 + 40x \\ 0 & = 4x^2 + 40x - 12 \phantom{.} 000 \\ 0 & = x^2 + 10x - 3000 \phantom{00} \text{ (Shown)} \end{align}
(d)
\begin{align} 0 & = x^2 + 10x - 3000 \\ 0 & = (x - 50)(x + 60) \end{align} \begin{align} x - 50 & = 0 && \text{ or } & x + 60 & = 0 \\ x & = 50 &&& x & = -60 \end{align}
(e) Use x = 50 instead of x = -60 since rate of water delivery cannot be negative
\begin{align} \text{From (b), time taken} & = {1200 \over x + 10} \\ & = {1200 \over 50 + 10} \\ & = 20 \text{ mins} \end{align}
Question 17
(a)
| x | 0 | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|---|
| y | -225 | 175 | 375 | 375 | 175 | -225 |
(b)(i)
\begin{align} \text{Max. point: } & (25, 400) \\ \\ \text{Max. profit } (y) & = \$ 400 \\ \\ \text{Selling price } (x) & = \$ 25 \end{align}
(b)(ii) No profit or loss means y = 0
\begin{align} \text{From graph, selling prices} & = \$ 5 \text{ or } \$ 45 \end{align}
(b)(iii)
\begin{align} \text{From graph, selling prices} & = \$ 20 \text{ or } \$ 30 \end{align}
Question 18
(a)(i)
\begin{align} \text{Radius of large circle} & = {12 \over 2} \\ & = 6 \text{ cm} \\ \\ \text{Radius of small circle} & = x \text{ cm} \\ \\ PQ & = (6 + x) \text{ cm} \end{align}
(a)(ii)
\begin{align} PS & = 6 \text{ cm} \\ \\ TS & = QR = x \text{ cm} \\ \\ PT & = (6 - x) \text { cm} \end{align}
(a)(iii)
\begin{align} DS & = \text{Radius of large circle} \\ & = 6 \text{ cm} \\ \\ RC & = \text{Radius of small circle} \\ & = x \text{ cm} \\ \\ SR & = 18 - 6 - x \\ & = (12 - x) \text{ cm} \end{align}
(b)
\begin{align} TQ & = SR = (12 - x) \text{ cm} \\ \\ \text{By Pytha} & \text{goras theorem, } \\ PQ^2 & = PT^2 + TQ^2 \\ (6 + x)^2 & = (6 - x)^2 + (12 - x)^2 \\ (6 + x)(6 + x) & = (6 - x)(6 - x) + (12 - x)(12 - x) \\ 36 + 6x + 6x + x^2 & = 36 - 6x - 6x + x^2 + 144 - 12x - 12x + x^2 \\ x^2 + 12x + 36 & = 2x^2 - 36x + 180 \\ 0 & = 2x^2 - x^2 - 36x - 12x + 180 - 36 \\ 0 & = x^2 - 48x + 144 \phantom{00} \text{ (Shown)} \end{align}
(c)
\begin{align} 0 & = x^2 - 48x + 144 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-48) \pm \sqrt{(-48)^2 - 4(1)(144)} \over 2(1)} \\ & = {48 \pm \sqrt{1728} \over 2} \\ & = 44.7846 \text{ or } 3.2154 \\ \\ [\text{Do } & \text{not use } x = 44.7846 \text{ since it is longer than the sides of the rectangle}] \\ \\ \text{Radius of small circle} & = x \\ & = 3.2154 \\ & \approx 3.22 \text{ cm} \end{align}
What’s next
- Chapter 2: More About Linear Inequalities
- All chapters of New Discovering Mathematics 3A & 3B
- O Level E Maths past year solutions — free, 2016 to the latest paper