(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}

(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}

(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}

(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}

(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}

(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}

(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 solutions} \end{align}

(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 solutions} \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 solutions} \end{align}

(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 solutions} \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}

(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}

(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}
$$ \text{Other root is } x = {1 \over 3} $$

(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 \\ (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}
$$ \text{For } k = 3, \text{ other root is } x = {3 \over 2} $$

(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 solutions} \end{align}

(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}

(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}

\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}
$$ \text{Her solution is incomplete as the root } x = 0 \text{ is eliminated} $$

\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}
$$ \text{His solution is incomplete as he left out the equation } 5x - 2 = - (2x + 1) $$

\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}