(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}

(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}

(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}

(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}

(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}

\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}

\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}

\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}

\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 solutions} \end{align}

(a)

(b)

(c) Look for the x-intercepts of the curve

$$ x = -2.55, 1.55 $$

(a)

(b)

(c) Look for the x-intercept of the curve

$$ x = - 2 $$

(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)

(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 } (-0.1, -2) \\ \\ \therefore k & < -2 \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

$$ \text{From graph, } x = -2.1, 0.8 $$

\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}

\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}

(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}

\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 } -4.6457 \\ & \approx 0.646 \text{ or } -4.65 \end{align}

\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 } -19.295 \\ \\ \text{Width of path} & = x \\ & = 1.2956 \\ & \approx 1.30 \text{ m} \end{align}

(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} \\ & = {2 \over 3} \text{ hours} \\ \\ {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}

(a)

t	0	0.5	1	1.5	2	2.5	3
y	15	14	12	9	5	0	-6

(b)

$$ \text{No. For } t \ge 2.5, y \le 0 $$

(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{Subsequently, the parcel is not moving} \end{align}

(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}

(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}