(a)

$$\begin{align} x^2 + 8x + 5 & = 0 \\ x^2 + 8x + \left(8 \over 2\right)^2 - \left(8 \over 2\right)^2 + 5 & = 0 \\ x^2 + 8x + 4^2 - 4^2 + 5 & = 0 \\ (x + 4)^2 - 16 + 5 & = 0 \\ (x + 4)^2 - 11 & = 0 \\ (x + 4)^2 & = 11 \\ x + 4 & = \pm \sqrt{11} \end{align}$$ $$\begin{align} x + 4 & = \sqrt{11} && \text{ or } & x + 4 & = - \sqrt{11} \\ x & = \sqrt{11} - 4 &&& x & = - \sqrt{11} - 4 \\ x & \approx -0.683 &&& x & \approx -7.32 \end{align}$$

(b)

$$\begin{align} x^2 + 7x - 3 & = 0 \\ x^2 + 7x + \left(7 \over 2\right)^2 - \left(7 \over 2\right)^2 - 3 & = 0 \\ x^2 + 7x + 3.5^2 - 3.5^2 - 3 & = 0 \\ (x + 3.5)^2 - 12.25 - 3 & = 0 \\ (x + 3.5)^2 - 15.25 & = 0 \\ (x + 3.5)^2 & = 15.25 \\ x + 3.5 & = \pm \sqrt{15.25} \end{align}$$ $$\begin{align} x + 3.5 & = \sqrt{15.25} && \text{ or } & x + 3.5 & = - \sqrt{15.25} \\ x & = \sqrt{15.25} - 3.5 &&& x & = - \sqrt{15.25} - 3.5 \\ x & \approx 0.405 &&& x & \approx -7.41 \end{align}$$

(c)

$$\begin{align} x^2 - 11x - 7 & = 0 \\ x^2 - 11x + \left(11 \over 2\right)^2 - \left(11 \over 2\right)^2 - 7 & = 0 \\ x^2 - 11x + 5.5^2 - 5.5^2 - 7 & = 0 \\ (x - 5.5)^2 - 30.25 - 7 & = 0 \\ (x - 5.5)^2 - 37.25 & = 0 \\ (x - 5.5)^2 & = 37.25 \\ x - 5.5 & = \pm \sqrt{37.25} \end{align}$$ $$\begin{align} x - 5.5 & = \sqrt{37.25} && \text{ or } & x - 5.5 & = - \sqrt{37.25} \\ x & = \sqrt{37.25} + 5.5 &&& x & = - \sqrt{37.25} + 5.5 \\ x & \approx 11.6 &&& x & \approx -0.603 \end{align}$$

(d)

$$\begin{align} x^2 + 1.2x & = 1 \\ x^2 + 1.2x + \left(1.2 \over 2\right)^2 - \left(1.2 \over 2\right)^2 & = 1 \\ x^2 + 1.2x + 0.6^2 - 0.6^2 & = 1 \\ (x + 0.6)^2 - 0.36 & = 1 \\ (x + 0.6)^2 & = 1 + 0.36 \\ (x + 0.6)^2 & = 1.36 \\ x + 0.6 & = \pm \sqrt{1.36} \end{align}$$ $$\begin{align} x + 0.6 & = \sqrt{1.36} && \text{ or } & x + 0.6 & = - \sqrt{1.36} \\ x & = \sqrt{1.36} - 0.6 &&& x & = -\sqrt{1.36} - 0.6 \\ x & \approx 0.566 &&& x & \approx -1.77 \end{align}$$

(a)

$$\begin{align} 2x^2 & + 6x + 1 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-6 \pm \sqrt{(6)^2 - 4(2)(1)} \over 2(2)} \\ & = {-6 \pm \sqrt{28} \over 4} \\ & = -0.1771 \text{ or } -2.822 \\ & \approx -0.177 \text{ or } -2.82 \end{align}$$

(b)

$$\begin{align} 3x^2 & - 7x - 2 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-7) \pm \sqrt{(-7)^2 - 4(3)(-2)} \over 2(3)} \\ & = {7 \pm \sqrt{73} \over 6} \\ & = 2.590 \text{ or } -0.2573 \\ & \approx 2.59 \text{ or } -0.257 \end{align}$$

(c)

$$\begin{align} -4x^2 & + x + 5 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-1 \pm \sqrt{(1)^2 - 4(-4)(5)} \over 2(-4)} \\ & = {-1 \pm \sqrt{81} \over -8} \\ & = -1 \text{ or } {5 \over 4} \end{align}$$

(d)

$$\begin{align} 3x^2 & = 5x + 1 \\ 3x^2 - 5x - 1 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-5) \pm \sqrt{(-5)^2 - 4(3)(-1)} \over 2(3)} \\ & = {5 \pm \sqrt{37} \over 6} \\ & = 1.847 \text{ or } -0.1804 \\ & \approx 1.85 \text{ or } - 0.180 \end{align}$$

(a)

$$\begin{align} (x - 3)^2 & = {4 \over 25} \\ x - 3 & = \pm \sqrt{4 \over 25} \\ x - 3 & = \pm {2 \over 5} \end{align}$$ $$\begin{align} x - 3 & = {2 \over 5} && \text{ or } & x - 3 & = -{2 \over 5} \\ x & = {2 \over 5} + 3 &&& x & = -{2 \over 5} + 3 \\ x & = {17 \over 5} &&& x & = {13 \over 5} \end{align}$$

(b)

$$\begin{align} (4 - x)^2 & = 12 \\ 4 - x & = \pm \sqrt{12} \end{align}$$ $$\begin{align} 4 - x & = \sqrt{12} && \text{ or } & 4 - x & = -\sqrt{12} \\ 4 - \sqrt{12} & = x &&& 4 + \sqrt{12} & = x \\ 0.536 & \approx x &&& 7.46 & \approx x \end{align}$$

(c)

$$\begin{align} (x - 1)(x + 3) & = 9 \\ x^2 + 3x - x - 3 & = 9 \\ x^2 + 2x - 3 & = 9 \\ x^2 + 2x - 3 - 9 & = 0 \\ x^2 + 2x - 12 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-2 \pm \sqrt{(2)^2 - 4(1)(-12)} \over 2(1)} \\ & = {-2 \pm \sqrt{52} \over 2} \\ & = 2.605 \text{ or } -4.605 \\ & \approx 2.61 \text{ or } -4.61 \end{align}$$

(d)

$$\begin{align} x(x + 4) & = 17 \\ x^2 + 4x & = 17 \\ x^2 + 4x - 17 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-4 \pm \sqrt{(4)^2 - 4(1)(-17)} \over 2(1)} \\ & = {-4 \pm \sqrt{84} \over 2} \\ & = 2.582 \text{ or } -6.582 \\ & \approx 2.58 \text{ or } -6.58 \end{align}$$

(i)

$$\begin{align} 2x^2 & - 7x + 4 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-7) \pm \sqrt{(-7)^2 - 4(2)(4)} \over 2(2)} \\ & = {7 \pm \sqrt{ 17} \over 4} \\ & = 2.780 \text{ or } 0.7192 \\ & \approx 2.78 \text{ or } 0.72 \text{ (2 d.p.)} \end{align}$$

(ii)

$$\begin{align} 2x^2 & - 7x + 4 = 0 \\ \\ \text{Replace } & x \text{ by } y - 1, \\ 2(y - 1)^2 & - 7(y - 1) + 4 = 0 \\ \\ \\ \text{From (i), } x & = 2.780 \text{ or } 0.7192 \\ y - 1 & = 2.780 \text{ or } 0.7192 \end{align}$$ $$\begin{align} y - 1 & = 2.780 && \text{ or } & y - 1 & = 0.7192 \\ y & = 2.780 + 1 &&& y & = 0.7192 + 1 \\ y & = 3.780 &&& y & = 1.7192 \\ y & \approx 3.78 &&& y & \approx 1.72 \end{align}$$

(a)

$$\begin{align} x & = 2 && \text{ or } & x & = {6 \over 7} \\ x - 2 & = 0 &&& 7x & = 6 \\ & &&& 7x - 6 & = 0 \end{align}$$ $$\begin{align} (x - 2)(7x - 6) & = 0 \\ 7x^2 - 6x - 14x + 12 & = 0 \\ 7x^2 - 20x + 12 & = 0 \end{align}$$

(b)

$$\begin{align} x & = -{1 \over 2} && \text{ or } & x & = -{2 \over 3} \\ 2x & = -1 &&& 3x & = -2 \\ 2x + 1 & =0 &&& 3x + 2 & = 0 \end{align}$$ $$\begin{align} (2x + 1)(3x + 2) & = 0 \\ 6x^2 + 4x + 3x + 2 & = 0 \\ 6x^2 + 7x + 2 & = 0 \end{align}$$

(a)

$$\begin{align} {x - 1 \over 1} & = {5 \over x + 7} \\ (x + 7)(x - 1) & = 5 \\ x^2 - x + 7x - 7 & = 5 \\ x^2 + 6x - 7 & = 5 \\ x^2 + 6x - 7 - 5 & = 0 \\ x^2 + 6x - 12 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-6 \pm \sqrt{(6)^2 - 4(1)(-12)} \over 2(1)} \\ & = {-6 \pm \sqrt{84} \over 2} \\ & = 1.582 \text{ or } -7.582 \\ & \approx 1.58 \text{ or } -7.58 \end{align}$$

(b)

$$\begin{align} {x - 1 \over x + 4} & = {2x \over x - 3} \\ (x - 3)(x - 1) & = 2x(x + 4) \\ x^2 - x - 3x + 3 & = 2x^2 + 8x \\ x^2 - 4x + 3 & = 2x^2 + 8x \\ 0 & = 2x^2 - x^2 + 8x + 4x - 3 \\ 0 & = x^2 + 12x - 3 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-12 \pm \sqrt{(12)^2 - 4(1)(-3)} \over 2(1)} \\ & = {-12 \pm \sqrt{156} \over 2} \\ & = 0.2449 \text{ or } -12.24 \\ & \approx 0.245 \text{ or } - 12.2 \end{align}$$

(c)

$$\begin{align} {1 \over x} - 5x & = 5 \\ {1 \over x} & = 5 + 5x \\ 1 & = x(5 + 5x) \\ 1 & = 5x + 5x^2 \\ 0 & = 5x^2 + 5x - 1 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-5 \pm \sqrt{(5)^2 - 4(5)(-1)} \over 2(5)} \\ & = {-5 \pm \sqrt{45} \over 10} \\ & = 0.1708 \text{ or } -1.170 \\ & \approx 0.171 \text{ or } -1.17 \end{align}$$

(d)

$$\begin{align} {5 \over x} & = 3 - {x \over x - 3} \\ {5 \over x} & = {3 \over 1} - {x \over x - 3} \\ {5 \over x} & = {3(x - 3) \over x - 3} - {x \over x - 3} \\ {5 \over x} & = {3(x - 3) - x \over x - 3} \\ {5 \over x} & = {3x - 9 - x \over x - 3} \\ {5 \over x} & = {2x - 9 \over x - 3} \\ 5(x - 3) & = x(2x - 9) \\ 5x - 15 & = 2x^2 - 9x \\ 0 & = 2x^2 - 9x - 5x + 15 \\ 0 & = 2x^2 - 14x + 15 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-14) \pm \sqrt{(-14)^2 - 4(2)(15)} \over 2(2)} \\ & = {14 \pm \sqrt{ 76 } \over 4 } \\ & = 5.679 \text{ or } 1.320 \\ & \approx 5.68 \text{ or } 1.32 \end{align}$$

(e)

$$\begin{align} {2 \over x + 1} + {1 \over x - 3} & = 5 \\ {2(x - 3) \over (x + 1)(x - 3)} + {x + 1 \over (x + 1)(x - 3)} & = 5 \\ {2(x - 3) + x + 1 \over (x + 1)(x - 3)} & = 5 \\ {2x - 6 + x + 1 \over x^2 - 3x + x - 3 } & = 5 \\ {3x - 5 \over x^2 - 2x - 3} & = {5 \over 1} \\ 3x - 5 & = 5(x^2 - 2x - 3) \\ 3x - 5 & = 5x^2 - 10x - 15 \\ 0 & = 5x^2 - 10x - 3x - 15 + 5 \\ 0 & = 5x^2 - 13x - 10 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-13) \pm \sqrt{(-13)^2 - 4(5)(-10)} \over 2(5)} \\ & = {13 \pm \sqrt{369} \over 10} \\ & = 3.221 \text{ or } -0.6209 \\ & \approx 3.22 \text{ or } -0.621 \end{align}$$

(f)

$$\begin{align} {x \over x + 1} + {1 \over 5} & = {3 \over x - 2} \\ {5x \over 5(x + 1)} + {x + 1 \over 5(x + 1)} & = {3 \over x - 2} \\ {5x + x + 1 \over 5(x + 1)} & = {3 \over x - 2} \\ {6x + 1 \over 5x + 5} & = {3 \over x - 2} \\ (x - 2)(6x + 1) & = 3(5x + 5) \\ 6x^2 + x - 12x - 2 & = 15x + 15 \\ 6x^2 - 11x - 2 & = 15x + 15 \\ 6x^2 - 11x - 15x - 2 - 15 & = 0 \\ 6x^2 - 26x - 17 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-26) \pm \sqrt{(-26)^2 - 4(6)(-17)} \over 2(6)} \\ & = {26 \pm \sqrt{1084} \over 12} \\ & = 4.910 \text{ or } 0.5770 \\ & \approx 4.91 \text{ or } 0.577 \end{align}$$

(g)

$$\begin{align} {5 \over x - 2} - {3 \over x^2 - 4} & = {2 \over 7} \\ {5 \over x - 2} - {3 \over \underbrace{(x + 2)(x - 2)}_{a^2 - b^2 = (a + b)(a - b)}} & = {2 \over 7} \\ {5(x + 2) \over (x + 2)(x - 2)} - {3 \over (x + 2)(x - 2)} & = {2 \over 7} \\ {5(x + 2) - 3 \over (x + 2)(x - 2)} & = {2 \over 7} \\ {5x + 10 - 3 \over x^2 - 4} & = {2 \over 7} \\ {5x + 7 \over x^2 - 4} & = {2 \over 7} \\ 7(5x + 7) & = 2(x^2 - 4) \\ 35x + 49 & = 2x^2 - 8 \\ 0 & = 2x^2 - 35x - 8 - 49 \\ 0 & = 2x^2 - 35x - 57 \\ 0 & = (x - 19)(2x + 3) \end{align}$$ $$\begin{align} x - 19 & = 0 && \text{ or } & 2x + 3 & = 0 \\ x & = 19 &&& 2x & = -3 \\ & &&& x & = -{3 \over 2} \end{align}$$

(h)

$$\begin{align} {1 \over 2x + 1} + {x + 3 \over 2x^2 - 5x - 3} & = 2 \\ {1 \over 2x + 1} + {x + 3 \over (2x + 1)(x - 3)} & = 2 \\ {x - 3 \over (2x + 1)(x - 3)} + {x + 3 \over (2x + 1)(x - 3)} & = 2 \\ {x - 3 + x + 3 \over (2x + 1)(x - 3)} & = 2 \\ {2x \over 2x^2 - 5x - 3} & = {2 \over 1} \\ 2x & = 2(2x^2 - 5x - 3) \\ 2x & = 4x^2 - 10x - 6 \\ 0 & = 4x^2 - 10x - 2x - 6 \\ 0 & = 4x^2 - 12x - 6 \\ 0 & = 2x^2 - 6x - 3 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-6) \pm \sqrt{(-6)^2 - 4(2)(-3)} \over 2(2)} \\ & = {6 \pm \sqrt{60} \over 4} \\ & = 3.436 \text{ or } -0.4364 \\ & \approx 3.44 \text{ or } -0.436 \end{align}$$

(i)

$$\begin{align} y & = x^2 - 7x + 12 \\ & = (x - 3)(x - 4) \end{align}$$

(ii)

$$\begin{align} y & = x^2 - 7x + 12 \phantom{000000} [\text{Minimum curve } \cup] \\ \\ \text{Let } & x = 0, \\ y & = (0)^2 - 7(0) + 12 \\ y & = 12 \phantom{000000000000000.} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = x^2 - 7x + 12 \\ 0 & = (x - 3)(x - 4) \\ \\ x & = 3 \phantom{0} \text{ or } \phantom{0} x = 4 \phantom{0000000} [x \text{-intercepts}] \\ \\ \text{Line of symmetry, } x & = {3 + 4 \over 2} \\ x & = 3.5 \\ \\ \text{Let } & x = 3.5, \\ y & = (3.5)^2 - 7(3.5) + 12 \\ y & = -0.25 \\ \\ \text{Minimum point: } & (3.5, -0.25) \end{align}$$

(i)

$$\begin{align} y & = - x^2 + 5x - 4 \\ & = - (x^2 - 5x) - 4 \\ & = - \left[ x^2 - 5x + \left(5 \over 2\right)^2 - \left(5 \over 2\right)^2 \right] - 4 \phantom{000000} [\text{Complete the square}] \\ & = - (x^2 - 5x + 2.5^2 - 2.5^2) - 4 \\ & = - [(x - 2.5)^2 - 6.25] - 4 \\ & = - (x - 2.5)^2 + 6.25 - 4 \\ & = - (x - 2.5)^2 + 2.25 \end{align}$$

(ii)

$$\begin{align} y & = - x^2 + 5x - 4 \phantom{000000000} [\text{Maximum curve } \cap] \\ y & = - (x - 2.5)^2 + 2.25 \\ \\ \text{Maximum point: } & (2.5, 2.25) \\ \\ \text{Let } & x = 0, \\ y & = - (0 - 2.5)^2 + 2.25 \\ y & = -4 \phantom{000000000000000000} [y \text{-intercept}] \end{align}$$

(i)

$$\begin{align} \text{In November, US\$} x & \rightarrow \text{£} 100 \\ \text{US\$} 1 & \rightarrow \text{£} {100 \over x} \\ \text{US\$} 650 & \rightarrow \text{£} {65 \phantom{.} 000 \over x} \\ \\ \\ \text{In December, US\$} (x - 5) & \rightarrow \text{£} 100 \\ \text{US\$} 1 & \rightarrow \text{£} {100 \over x - 5} \\ \text{US\$} 650 & \rightarrow \text{£} {65 \phantom{.} 000 \over x - 5} \\ \\ \\ {65 \phantom{.} 000 \over x - 5} - {65 \phantom{.} 000 \over x} & = 20 \end{align}$$

(ii)

$$\begin{align} {65 \phantom{.} 000 \over x - 5} - {65 \phantom{.} 000 \over x} & = 20 \\ {65 \phantom{.} 000 x \over x(x - 5)} - {65 \phantom{.} 000 (x - 5) \over x(x - 5)} & = 20 \\ {65 \phantom{.} 000x - 65 \phantom{.} 000(x - 5) \over x(x - 5)} & = 20 \\ {65 \phantom{.} 000x - 65 \phantom{.} 000x + 325 \phantom{.} 000 \over x^2 - 5x} & = 20 \\ {325 \phantom{.} 000 \over x^2 - 5x} & = {20 \over 1} \\ 325 \phantom{.} 000 & = 20(x^2 - 5x) \\ {325 \phantom{.} 000 \over 20} & = x^2 - 5x \\ 16 \phantom{.} 250 & = x^2 - 5x \\ 0 & = x^2 - 5x - 16 \phantom{.} 250 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-5) \pm \sqrt{(-5)^2 - 4(1)(-16 \phantom{.} 250)} \over 2(1)} \\ & = {5 \pm \sqrt{65 \phantom{.} 025} \over 2} \\ & = 130 \text{ or } -125 \\ \\ \\ \text{In November, £} 100 & \rightarrow \text{US\$} x \\ \text{£} 1250 & \rightarrow \text{US\$} 12.5x \\ \\ \text{Amount of dollars received} & = 12.5x \\ & = 12.5 (130) \\ & = \text{US\$} 1625 \end{align}$$

(i)

$$\begin{align} \text{Time} & = { \text{Distance} \over \text{Speed} } \\ \\ \text{Time taken by Ali} & = {40 \over x} \text{ hours} \\ \\ \text{Time taken by Kumar} & = {40 \over x - 30} \text{ hours} \end{align}$$

(ii)

$$\begin{align} 10 \text{ mins} & = {1 \over 6} \text{ hours} \\ \\ {40 \over x} + {1 \over 6} & = {40 \over x - 30} \\ {6(40) \over 6x} + {x \over 6x} & = {40 \over x - 30} \\ {240 + x \over 6x} & = {40 \over x - 30} \\ (x - 30)(240 + x) & = 40(6x) \\ 240x + x^2 - 7200 - 30x & = 240x \\ x^2 + 210x - 7200 & = 240x \\ x^2 + 210x - 240x - 7200 & = 0 \\ x^2 - 30x - 7200 & = 0 \phantom{00} \text{ (Shown)} \end{align}$$

(iii)

$$\begin{align} x^2 & - 30x - 7200 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-30) \pm \sqrt{(-30)^2 - 4(1)(-7200)} \over 2(1)} \\ & = {30 \pm \sqrt{29 \phantom{.} 700} \over 2} \\ & = 101.168 \text{ or } -71.168 \\ & \approx 101.17 \text{ or } -71.17 \text{ (2 d.p.)} \end{align}$$

(iv)

$$\begin{align} \text{Since Ali's average speed is greater than 0 km/h, reject } x = -71.168 \end{align}$$

(v)

$$\begin{align} \text{Time taken by Kumar} & = {40 \over x - 30} \\ & = {40 \over 101.168 - 30} \\ & = 0.56205 \text{ hours} \\ & = (0.56205 \times 60) \text{ mins} \\ & \approx 33.7 \text{ mins} \end{align}$$

(i)

$$\begin{align} x \text{ g of sugar} & \rightarrow 1 \text{ cake} \\ 1 \text{ g of sugar} & \rightarrow {1 \over x} \text{ cake} \\ 1 \text{ kg of sugar} & \rightarrow {1000 \over x} \text{ cake} \phantom{000000} [\text{1 kg = 1000 g}] \\ 6 \text{ kg of sugar} & \rightarrow {6000 \over x} \text{ cakes} \end{align}$$

(ii)

$$\begin{align} (x + 10) \text{ g of sugar} & \rightarrow 1 \text{ cake} \\ 1 \text{ g of sugar} & \rightarrow {1 \over x + 10} \text{ cake} \\ 1 \text{ kg of sugar} & \rightarrow {1000 \over x + 10} \text{ cake} \\ 6 \text{ kg of sugar} & \rightarrow {6000 \over x + 10} \text{ cakes} \end{align}$$

(iii)

$$\begin{align} {6000 \over x} - {6000 \over x + 10} & = 3 \\ {6000(x + 10) \over x(x + 10)} - {6000x \over x(x + 10)} & = 3 \\ {6000(x + 10) - 6000x \over x(x + 10)} & = 3 \\ {6000x + 60 \phantom{.} 000 - 6000x \over x^2 + 10x} & = 3 \\ {60 \phantom{.} 000 \over x^2 + 10x} & = {3 \over 1} \\ 60 \phantom{.} 000 & = 3(x^2 + 10x) \\ {60 \phantom{.} 000 \over 3} & = x^2 + 10x \\ 20 \phantom{.} 000 & = x^2 + 10x \\ 0 & = x^2 + 10x - 20 \phantom{.} 000 \phantom{00} \text{ (Shown)} \end{align}$$

(iv)

$$\begin{align} 0 & = x^2 + 10x - 20 \phantom{.} 000 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-10 \pm \sqrt{(10)^2 - 4(1)(-20 \phantom{.} 000)} \over 2(1)} \\ & = {-10 \pm \sqrt{80 \phantom{.} 100} \over 2} \\ & = 136.51 \text{ or } -146.51 \\ & \approx 137 \text{ or } -147 \end{align}$$

(v)

$$\begin{align} \text{Cakes baked by Devi} & = {6000 \over x} \\ & = {6000 \over 136.51} \\ & = 43.952 \\ & \approx 43 \phantom{000000} [\text{Round down}] \\ \\ \text{Cakes baked by Siti} & = {6000 \over x + 10} \\ & = {6000 \over 136.51 + 10} \\ & = 40.952 \\ & \approx 40 \\ \\ \text{Money collected} & = (43 + 40) \times 15 \\ & = \$ 1245 \end{align}$$

(i)

$$\begin{align} \text{Length of floor} & = (35 - 2x) \text{ m} \\ \\ \text{Breadth of floor} & = (22 - 2x) \text{ m} \end{align}$$

(ii)

$$\begin{align} \text{Floor area} & = (35 - 2x)(22 - 2x) \\ 400 & = 770 - 70x - 44x + 4x^2 \\ 400 & = 4x^2 - 114x + 770 \\ 0 & = 4x^2 - 114x + 770 - 400 \\ 0 & = 4x^2 - 114x + 370 \\ 0 & = 2x^2 - 57x + 185 \phantom{00} \text{ (Shown)} \end{align}$$

(iii)

$$\begin{align} 0 & = 2x^2 - 57x + 185 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-57) \pm \sqrt{(-57)^2 - 4(2)(185)} \over 2(2)} \\ & = {57 \pm \sqrt{1769} \over 4} \\ & = 24.764 \text{ or } 3.735 \\ & \approx 24.76 \text{ or } 3.74 \text{ (2 d.p.)} \end{align}$$

(iv) Reject x = 24.764 since it is longer than the width of the room (22 m)

$$\begin{align} \text{Width of floor} & = x \\ & \approx 3.74 \text{ m} \end{align}$$

(i)

$$\begin{align} 60 + 25x - x^2 & = 0 \\ - x^2 + 25x + 60 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-25 \pm \sqrt{(25)^2 - 4(-1)(60)} \over 2(-1)} \\ & = {-25 \pm \sqrt{865} \over -2} \\ & = -2.205 \text{ or } 27.205 \\ & \approx -2.2 \text{ or } 27.2 \text{ (1 d.p.)} \\ \\ \\ x \approx 27.2 \text{ represents } & \text{the horizontal distance from the tower} \\ \text{when the stone is at } & \text{sea level } (y = 0) \end{align}$$

(ii)

(iii)(a)

$$\begin{align} \text{Greatest height } (y) & = 216 \text{ m} \end{align}$$

(iii)(b)

$$\begin{align} \text{Horizontal distance } (x) & = 6.5 \text{ m} \text{ or } 18.5 \text{ m} \end{align}$$

$$\begin{align} \text{Let the time taken by the} & \text{ larger pipe be } x \text{ mins} \\ \\ x \text{ mins} & \rightarrow 1 \text{ tank} \\ 1 \text{ min} & \rightarrow {1 \over x} \text{ tank} \\ 11{1 \over 9} \text{ mins} & \rightarrow {100 \over 9x} \text{ tank} \phantom{000000} \left[ 11{1 \over 9} = {100 \over 9} \right] \\ \\ \\ \text{Time taken by smaller pipe} & = (x + 5) \text{ mins} \\ \\ (x + 5) \text{ mins} & \rightarrow 1 \text{ tank} \\ 1 \text{ min} & \rightarrow {1 \over x + 5} \text{ tank} \\ 11{1 \over 9} \text{ mins} & \rightarrow {100 \over 9(x + 5)} \text{ tank} \\ \\ \\ {100 \over 9x} + {100 \over 9(x + 5)} & = 1 \\ {100(x + 5) \over 9x(x + 5)} + {100x \over 9(x + 5)} & = 1 \\ {100(x + 5) + 100x \over 9x(x + 5)} & = 1 \\ {100x + 500 + 100x \over 9x^2 + 45x} & = 1 \\ {200x + 500 \over 9x^2 + 45x} & = {1 \over 1} \\ 200x + 500 & = 9x^2 + 45x \\ 0 & = 9x^2 + 45x - 200x - 500 \\ 0 & = 9x^2 - 155x - 500 \\ 0 & = (x - 20)(9x + 25) \end{align}$$ $$\begin{align} x - 20 & = 0 && \text{ or } & 9x + 25 & = 0 \\ x & = 20 &&& 9x & = -25 \\ & &&& x & = -{25 \over 9} \text{ (Reject, since } x > 0) \end{align}$$
$$\begin{align} \text{Time taken by larger pipe} & = x \\ & = 20 \text{ mins} \\ \\ \text{Time taken by smaller pipe} & = x + 5 \\ & = 20 + 5 \\ & = 25 \text{ mins} \end{align}$$

$$\begin{align} \text{Time} & = { \text{Distance} \over \text{Speed} } \\ \\ \text{Time taken (against current)} & = {12 \over x - 5} \text{ hours} \\ \\ \text{Time taken (with current)} & = {12 \over x + 5} \text{ hours} \\ \\ 1 \text{ hour } 30 \text{ mins} & = 1{1 \over 2} \text{ hours} \\ \\ {12 \over x + 5} + {12 \over x - 5} & = {3 \over 2} \\ {12(x - 5) \over (x + 5)(x - 5)} + {12(x + 5) \over (x + 5)(x - 5)} & = {3 \over 2} \\ {12(x - 5) + 12(x + 5) \over (x + 5)(x - 5)} & = {3 \over 2} \\ {12x - 60 + 12x + 60 \over \underbrace{x^2 - 5^2}_{(a + b)(a - b) = a^2 - b^2}} & = {3 \over 2} \\ {24x \over x^2 - 25} & = {3 \over 2} \\ 2(24x) & = 3(x^2 - 25) \\ 48x & = 3x^2 - 75 \\ 0 & = 3x^2 - 48x - 75 \\ 0 & = x^2 - 16x - 25 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-16) \pm \sqrt{(-16)^2 - 4(1)(-25)} \over 2(1)} \\ & = {16 \pm \sqrt{356} \over 2} \\ & = 17.434 \text{ or } -1.4339 \text{ (Reject, since } x > 0) \\ \\ \\ \text{Speed of boat in still water} & = x \\ & = 17.434 \\ & \approx 17.4 \text{ km/h} \end{align}$$