(a)

\begin{align*} 4x^2 + 9x + 1 & = 0 \\ x^2 + 2.25x + 0.25 & = 0 \\ \left(x + {2.25 \over 2}\right)^2 - \left(2.25 \over 2\right)^2 + 0.25 & = 0 \\ \left( x + {9 \over 8} \right)^2 - {81 \over 64} + 0.25 & = 0 \\ \left( x + {9 \over 8} \right)^2 - {65 \over 64} & = 0 \\ \left( x + {9 \over 8} \right)^2 & = {65 \over 64} \\ x + {9 \over 8} & = \pm \sqrt{65 \over 64} \\ x + {9 \over 8} & = \pm 1.0077 \\ x & = \pm 1.0077 - {9 \over 8} \\ x & = -0.11721 \text{ or } -2.1327 \\ x & \approx -0.117 \text{ or } -2.13 \end{align*}

(b)

\begin{align*} 8x^2 - 13x - 2 & = 0 \\ x^2 - {13 \over 8}x - {1 \over 4} & = 0 \\ \left(x - {13 \over 16}\right)^2 - \left(13 \over 16\right)^2 - {1 \over 4} & = 0 \\ \left(x - {13 \over 16}\right)^2 - {233 \over 256} & = 0 \\ \left(x - {13 \over 16}\right)^2 & = {233 \over 256} \\ x - {13 \over 16} & = \pm \sqrt{233 \over 256} \\ x - {13 \over 16} & = \pm 0.95402 \\ x & = \pm 0.95402 + {13 \over 16} \\ x & = 1.7665 \text{ or } -0.14152 \\ x & \approx 1.77 \text{ or } -0.142 \end{align*}

(c)

\begin{align*} 10x - 2 - 3x^2 & = 0 \\ -3x^2 + 10x - 2 & = 0 \\ x^2 - {10 \over 3}x + {2 \over 3} & = 0 \\ \left(x - {5 \over 3}\right)^2 - \left(5 \over 3\right)^2 + {2 \over 3} & = 0 \\ \left(x - {5 \over 3}\right)^2 - {19 \over 9} & = 0 \\ \left(x - {5 \over 3}\right)^2 & = {19 \over 9} \\ x - {5 \over 3} & = \pm \sqrt{19 \over 9} \\ x - {5 \over 3} & = \pm 1.4529 \\ x & = \pm 1.4529 + {5 \over 3} \\ x & = 3.1196 \text{ or } 0.2137 \\ x & \approx 3.12 \text{ or } 0.214 \end{align*}

(d)

\begin{align*} (6x + 1)(x + 1) & = 5 \\ 6x^2 + 6x + x + 1 & = 5 \\ 6x^2 + 7x + 1 & = 5 \\ 6x^2 + 7x - 4 & = 0 \\ x^2 + {7 \over 6}x - {2 \over 3} & = 0 \\ \left(x + {7 \over 12}\right)^2 - \left(7 \over 12\right)^2 - {2 \over 3} & = 0 \\ \left(x + {7 \over 12}\right)^2 - {145 \over 144} & = 0 \\ \left(x + {7 \over 12}\right)^2 & = {145 \over 144} \\ x + {7 \over 12} & = \pm \sqrt{145 \over 144} \\ x + {7 \over 12} & = \pm 1.0034 \\ x & = \pm 1.0034 - {7 \over 12} \\ x & = 0.42013 \text{ or } -1.5867 \\ x & \approx 0.420 \text{ or } -1.59 \end{align*}

(a)

\begin{align*} 3x^2 + x + 17 & = 0 \\ \\ [a = 3, b & = 1, c = 17] \\ \\ b^2 - 4ac & = (1)^2 - 4(3)(17) \\ & = -203 \\ \\ \text{Since } b^2 - 4ac < 0, & \text{ the equation has no real roots} \end{align*}

(b)

\begin{align*} 5x^2 - x + 8 & = 0 \\ \\ [a = 5, b & = -1, c = 8] \\ \\ b^2 - 4ac & = (-1)^2 - 4(5)(8) \\ & = -159 \\ \\ \text{Since } b^2 - 4ac < 0, & \text{ the equation has no real roots} \end{align*}

(c)

\begin{align*} -2x^2 + 9x & = 21 \\ -2x^2 + 9x - 21 & = 0 \\ \\ [a = -2, b & = 9, c = -21] \\ \\ b^2 - 4ac & = (9)^2 - 4(-2)(-21) \\ & = -87 \\ \\ \text{Since } b^2 - 4ac < 0, & \text{ the equation has no real roots} \end{align*}

(d)

\begin{align*} (7x + 1)(7x + 2) & = 25x - x^2 \\ 49x^2 + 14x + 7x + 2 & = 25x - x^2 \\ 49x^2 + 21x + 2 & = 25x - x^2 \\ 50x^2 - 4x + 2 & = 0 \\ 25x^2 - 2x + 1 & = 0 \\ \\ [a = 25, b & = -2, c = 1] \\ \\ b^2 - 4ac & = (-2)^2 - 4(25)(1) \\ & = -96 \\ \\ \text{Since } b^2 - 4ac < 0, & \text{ the equation has no real roots} \end{align*}

(a)

\begin{align*} 3x^2 - 12x + k & = 4 \\ 3x^2 - 12x + k - 4 & = 0 \\ \\ [a = 3, b & = -12, c = k - 4] \\ \\ b^2 - 4ac & = (-12)^2 - 4(3)(k - 4) \\ & = 144 - 12(k - 4) \\ & = 144 - 12k + 48 \\ & = 192 - 12k \\ \\ \\ b^2 - 4ac & \ge 0 \phantom{0000} [\text{Two real roots}] \\ 192 - 12k & \ge 0 \\ -12k & \ge -192 \\ 12k & \le 192 \\ k & \le {192 \over 12} \\ k & \le 16 \end{align*}

(b)

\begin{align*} b^2 - 4ac & < 0 \phantom{0000} [\text{No real roots}] \\ 192 - 12k & < 0 \\ -12k & < -192 \\ 12k & > 192 \\ k & > {192 \over 12} \\ k & > 16 \end{align*}

\begin{align*} x^2 + x(2x + p) + 4 & = 0 \\ x^2 + 2x^2 + px + 4 & = 0 \\ 3x^2 + px + 4 & = 0 \\ \\ [a = 3, b & = p, c = 4 ] \\ \\ b^2 - 4ac & = (p)^2 - 4(3)(4) \\ & = p^2 - 48 \\ \\ b^2 - 4ac & = 0 \phantom{00000} [\text{Two real and equal roots}] \\ p^2 - 48 & = 0 \\ p^2 & = 48 \\ p & = \pm {48} \\ p & = \pm \sqrt{16} \sqrt{3} \\ p & = \pm 4 \sqrt{3} \end{align*}

\begin{align*} kx^2 + 2kx - 4 + k & = 0 \\ \\ [a = k, b & = 2k, c = k - 4 ] \\ \\ b^2 - 4ac & = (2k)^2 - 4(k)(k - 4) \\ & = 4k^2 - 4k(k - 4) \\ & = 4k^2 - 4k^2 + 16k \\ & = 16k \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ 16k & < 0 \\ k & < {0 \over 16} \\ k & < 0 \end{align*}

\begin{align*} hx^2 + 8x + 3k & = 6 \\ hx^2 + 8x + 3k - 6 & = 0 \\ \\ [a = h, b & = 8, c = 3k - 6 ] \\ \\ b^2 - 4ac & = (8)^2 - 4(h)(3k - 6) \\ & = 64 - 4h(3k - 6) \\ & = 64 - 12hk + 24h \\ \\ b^2 - 4ac & = 0 \phantom{00000} [\text{Two real and equal roots}] \\ 64 - 12hk + 24h & = 0 \\ -12hk + 24h & = -64 \\ 12hk - 24h & = 64 \\ 3hk - 6h & = 16 \\ \\ \text{If } & h = 1, \\ 3(1)k - 6(1) & = 16 \\ 3k - 6 & = 16 \\ 3k & = 22 \\ k & = {22 \over 3} \\ \\ \therefore h & = 1, k = {22 \over 3} \end{align*}

\begin{align*} 2x^2 + 5cx & = 3c \\ 2x^2 + 5cx - 3c & = 0 \\ \\ [a = 2, b & = 5c, \text{ '}c\text{'} = -3c ] \\ \\ b^2 - 4ac & = (5c)^2 - 4(2)(-3c) \\ & = 25c^2 - 8(-3c) \\ & = 25c^2 + 24c \\ \\ \text{For positive} \text{ values of } & c, 25c^2 > 0 \text{ and } 24c > 0 \\ 25c^2 + 24c & > 0 \\ b^2 - 4ac & > 0 \\ \\ \therefore \text{The equation has real} & \text{ and distinct roots for all positive values of } c \end{align*}

(i)

\begin{align*} x^2 - 3b & = 1 - 3bx \\ x^2 + 3bx - 3b - 1 & = 0 \\ x^2 + 3bx + (-3b - 1) & = 0 \\ \\ [a = 1, b & = 3b, c = -3b - 1 ] \\ \\ b^2 - 4ac & = (3b)^2 - 4(1)(-3b - 1) \\ & = 9b^2 - 4(-3b - 1) \\ & = 9b^2 + 12b + 4 \\ & = (3b)^2 + 2(3b)(2) + (2)^2 \\ & = (3b + 2)^2 \phantom{0000000000000} [a^2 + 2ab + b^2 = (a + b)^2] \\ \\ \text{For all real values} & \text{ of } b, \\ (3b + 2)^2 & \ge 0 \\ b^2 - 4ac & \ge 0 \\ \\ \therefore \text{The equation has real} & \text{ roots for all real values of } b \end{align*}

(ii) If the equation has two real and distinct roots, b² - 4ac must be greater than 0. For this to happen, b can be any value except $-{2 \over 3}$

$$ b = 5 $$

\begin{align*} x^2 + 2kx & + k^2 - 5k + 7 \\ \\ [a = 1, b & = 2k, c = k^2 - 5k + 7 ] \\ \\ \text{Since } a > 0, & \text{ the expression has a minimum value} \\ \\ b^2 - 4ac & = (2k)^2 - 4(1)(k^2 - 5k + 7) \\ & = 4k^2 - 4(k^2 - 5k + 7) \\ & = 4k^2 - 4k^2 + 20k - 28 \\ & = 20k - 28 \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ 20k - 28 & < 0 \\ 20k & < 28 \\ k & < {28 \over 20} \\ k & < {7 \over 5} \end{align*}

\begin{align*} (2p + 3)x^2 & + (4p - 14)x + 16p + 1 \\ \\ [a = 2p + 3, b & = 4p - 14, c = 16p + 1 ] \\ \\ \text{For the expression } & \text{to be always negative, } \\ a & < 0 \\ 2p + 3 & < 0 \\ 2p & < -3 \\ p & < -{3 \over 2} \phantom{00000} [\text{Condition #1}] \\ \\ \\ b^2 - 4ac & = (4p - 14)^2 - 4(2p + 3)(16p + 1) \\ & = (4p)^2 - 2(4p)(14) + (14)^2 - 4(32p^2 + 2p + 48p + 3) \\ & = 16p^2 - 112p + 196 - 4(32p^2 + 50p + 3) \\ & = 16p^2 - 112p + 196 - 128p^2 - 200p - 12 \\ & = -112p^2 - 312p + 184 \end{align*}
\begin{align*} b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ -112p^2 - 312p + 184 & < 0 \\ 112p^2 + 312p - 184 & > 0 \\ 14p^2 + 39p - 23 & > 0 \\ (7p + 23)(2p - 1) & > 0 \end{align*}

$$ p < -3{2 \over 7} \text{ or } p > {1 \over 2} \phantom{00000} [\text{Condition #2}] $$
Note we need to satisfy both conditions #1 and #2
$$ \therefore p < - 3{2 \over 7} $$

(i)

\begin{align*} ax^2 & + 9x + c \\ \\ \text{For the expression } & \text{to be always negative, } \\ a & < 0 \phantom{00000} [\text{Condition #1}] \\ \\ \\ b^2 - 4ac & = (9)^2 - 4(a)(c) \\ & = 81 - 4ac \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ 81 - 4ac & < 0 \\ -4ac & < -81 \\ 4ac & > 81 \\ ac & > {81 \over 4} \phantom{00000} [\text{Condition #2}] \end{align*}

(ii)

\begin{align*} \text{If } & a = -1, \\ (-1)c & > {81 \over 4} \\ c & < -{81 \over 4} \\ c & < -20{1 \over 4} \\ \\ \therefore a & = -1, c = -21 \end{align*}

\begin{align*} x^2 + (1 - a)x & = a \\ x^2 + (1 - a)x - a & = 0 \\ \\ b^2 - 4ac & = (1 - a)^2 - 4(1)(-a) \\ & = (1)^2 - 2(1)(a) + (a)^2 - 4(-a) \\ & = 1 - 2a + a^2 + 4a \\ & = a^2 + 2a + 1 \\ & = (a)^2 + 2(a)(1) + (1)^2 \\ & = (a + 1)^2 \phantom{00000000000} [a^2 + 2ab + b^2 = (a + b)^2] \\ \\ \text{For all real values} & \text{ of } a, \\ (a + 1)^2 & \ge 0 \\ b^2 - 4ac & \ge 0 \\ \\ \therefore \text{The equation has real} & \text{ roots for all real values of } a \end{align*}

(i)

\begin{align*} px^2 - 3x - p & = 0 \\ \\ b^2 - 4ac & = (-3)^2 - 4(p)(-p) \\ & = 9 - 4p(-p) \\ & = 9 + 4p^2 \\ \\ \text{For all real} & \text{ values of } p, \\ 9 + 4p^2 & > 0 \\ \implies b^2 - 4ac & > 0 \\ \text{Equation has two } & \text{real and distinct roots for all real values of } p \end{align*}

(ii)

No. Since the equation $px^2 - 3x - p = 0$ has two real and distinct roots for all real values of $p$, the curve $y = px^2 - 3x - p$ will intersect the $x$-axis twice.