Think! Additional Mathematics Textbook & Workbook (10th edition) Solutions
Worksheet 2A
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Solutions
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(a)
\begin{align} 4x^2 + 9x + 1 & = 0 \\ 4 \left( x^2 + {9 \over 4}x \right) + 1 & = 0 \\ 4 \left[ x^2 + {9 \over 4x} + \left(9 \over 8\right)^2 - \left(9 \over 8\right)^2 \right] + 1 & = 0 \\ 4 \left[ \left(x + {9 \over 8}\right)^2 - {81 \over 64} \right] + 1 & = 0 \\ 4 \left(x + {9 \over 8}\right)^2 - {81 \over 16} + 1 & = 0 \\ 4 \left(x + {9 \over 8}\right)^2 - {65 \over 16} & = 0 \\ 4 \left(x + {9 \over 8}\right)^2 & = {65 \over 16} \\ \left(x + {9 \over 8}\right)^2 & = {65 \over 64} \\ x + {9 \over 8} & = \pm \sqrt{65 \over 64} \\ x & = \pm \sqrt{65 \over 64} - {9 \over 8} \\ x & = -0.11721 \text{ or } -2.1327 \\ x & \approx -0.12 \text{ or } -2.13 \text{ (2 d.p.)} \end{align}
(b)
\begin{align} 2 - 11x + 6x^2 & = 0 \\ 6x^2 - 11x + 2 & = 0 \\ 6 \left(x^2 - {11 \over 6} x \right) + 2 & = 0 \\ 6 \left[ x^2 - {11 \over 6}x + \left(11 \over 12\right)^2 - \left(11 \over 12\right)^2 \right] + 2 & = 0 \\ 6 \left[ \left(x - {11 \over 12}\right)^2 - {121 \over 144} \right] + 2 & = 0 \\ 6 \left(x - {11 \over 12}\right)^2 - {121 \over 24} + 2 & = 0 \\ 6 \left(x - {11 \over 12}\right)^2 - {73 \over 24} & = 0 \\ 6 \left(x - {11 \over 12}\right)^2 & = {73 \over 24} \\ \left(x - {11 \over 12}\right)^2 & = {73 \over 144} \\ x - {11 \over 12} & = \pm \sqrt{73 \over 144} \\ x & = \pm \sqrt{73 \over 144} + {11 \over 12} \\ x & = 1.6286 \text{ or } 0.20466 \\ x & \approx 1.63 \text{ or } 0.20 \text{ (2 d.p.)} \end{align}
(c)
\begin{align} -{1 \over 5}x^2 & = x - 8 \\ -5 \left(-{1 \over 5}x^2\right) & = -5(x - 8) \\ x^2 & = -5x + 40 \\ x^2 + 5x - 40 & = 0 \\ x^2 + 5x + \left(5 \over 2\right)^2 - \left(5 \over 2\right)^2 - 40 & = 0 \\ \left(x + {5 \over 2}\right)^2 - {25 \over 4} - 40 & = 0 \\ \left(x + {5 \over 2}\right)^2 - {185 \over 4} & = 0 \\ \left(x + {5 \over 2}\right)^2 & = {185 \over 4} \\ x + {5 \over 2} & = \pm \sqrt{185 \over 4} \\ x & = \pm \sqrt{185 \over 4} - {5 \over 2} \\ x & = 4.3007 \text{ or } -9.3007 \\ x & \approx 4.30 \text{ or } -9.30 \text{ (2 d.p.)} \end{align}
(d)
\begin{align} (3x + 1)(2x - 4) & = 7 \\ 6x^2 - 12x + 2x - 4 & = 7 \\ 6x^2 - 10x - 4 & = 7 \\ 6x^2 - 10x - 4 - 7 & = 0 \\ 6x^2 - 10x - 11 & = 0 \\ 6 \left( x^2 - {5 \over 3}x \right) - 11 & = 0 \\ 6 \left[ x^2 - {5 \over 3}x + \left(5 \over 6\right)^2 - \left(5 \over 6\right)^2 \right] - 11 & = 0 \\ 6 \left[ \left(x - {5 \over 6}\right)^2 - {25 \over 36} \right] - 11 & = 0 \\ 6 \left(x - {5 \over 6}\right)^2 - {25 \over 6} - 11 & = 0 \\ 6 \left(x - {5 \over 6}\right)^2 - {91 \over 6} & = 0 \\ 6 \left(x - {5 \over 6}\right)^2 & = {91 \over 6} \\ \left(x - {5 \over 6}\right)^2 & = {91 \over 36} \\ x - {5 \over 6} & = \pm \sqrt{91 \over 36} \\ x & = \pm \sqrt{91 \over 36} + {5 \over 6} \\ x & = 2.4232 \text{ or } -0.75656 \\ x & \approx 2.42 \text{ or } -0.76 \text{ (2 d.p.)} \end{align}
(a)
\begin{align} 5x^2 & - x - 12 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-1) \pm \sqrt{(-1)^2 - 4(5)(-12)} \over 2(5)} \\ & = {1 \pm \sqrt{241} \over 10} \\ & = 1.6524 \text{ or } -1.4524 \\ & \approx 1.65 \text{ or } -1.45 \end{align}
(b)
\begin{align} 7x & = 9 - 2x^2 \\ 2x^2 + 7x - 9 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-7 \pm \sqrt{(7)^2 - 4(2)(-9)} \over 2(2)} \\ & = {-7 \pm \sqrt{121} \over 2(2)} \\ & = 1 \text{ or } -4.5 \end{align}
(c)
\begin{align} 1.4x(x + 3) + 1 & = 0 \\ 1.4x^2 + 4.2x + 1 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-4.2 \pm \sqrt{(4.2)^2 - 4(1.4)(1)} \over 2(1.4)} \\ & = {-4.2 \pm \sqrt{12.04} \over 2.8} \\ & = -0.26076 \text{ or } -2.7392 \\ & \approx -0.261 \text{ or } -2.74 \end{align}
(d)
\begin{align} (6x - 5)(x + 4) & = 3x^2 \\ 6x^2 + 24x - 5x - 20 & = 3x^2 \\ 6x^2 + 19x - 20 & = 3x^2 \\ 3x^2 + 19x - 20 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-19 \pm \sqrt{(19)^2 - 4(3)(-20)} \over 2(3)} \\ & = {-19 \pm \sqrt{601} \over 6} \\ & = 0.91921 \text{ or } -7.2525 \\ & \approx 0.919 \text{ or } -7.25 \end{align}
\begin{align} x^2 - 9x + 22 & = x^2 - 9x + \left(9 \over 2\right)^2 - \left(9 \over 2\right)^2 + 22 \\ & = \left(x - {9 \over 2}\right)^2 - {81 \over 4} + 22 \\ & = \left(x - {9 \over 2}\right)^2 + {7 \over 4} \\ \\ \text{For all real} & \text{ values of } x, \\ \left(x - {9 \over 2}\right)^2 & \ge 0 \\ \left(x - {9 \over 2}\right)^2 + {7 \over 4} & \ge {7 \over 4} \\ \\ \therefore x^2 - 9x + 22 = 0 & \text{ has no solutions} \end{align}
(i)
\begin{align}
5x + 1 & = {18 \over x} \\
x(5x + 1) & = 18 \\
5x^2 + x & = 18 \\
5x^2 + x - 18 & = 0 \\
\\
(5x - 9)(x + 2) & = 0
\phantom{0000000} [\text{Factorise}]
\end{align}
\begin{align}
5x - 9 & = 0 && \text{ or } & x + 2 & = 0 \\
5x & = 9 &&& x & = -2 \\
x & = {9 \over 5} \\
x & = 1.8
\end{align}
\begin{align}
5x^2 & + x - 18 = 0 \\
\\
x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\
& = {-1 \pm \sqrt{(1)^2 - 4(5)(-18)} \over 2(5)} \\
& = {-1 \pm \sqrt{361} \over 10} \\
& = 1.8 \text{ or } -2
\end{align}
(ii)
\begin{align}
5x^2 + x - 18 & = 0
&&&
\text{Roots: } x & = 1.8, -2 \\
\\
\text{Replace } x & \text{ by } \sqrt{x}, \\
5(\sqrt{x})^2 + \sqrt{x} - 18 & = 0
&&&
\text{Roots: } \sqrt{x} & = 1.8, -2 \\
5x + \sqrt{x} - 18 & = 0
\end{align}
\begin{align}
\sqrt{x} & = 1.8
&& \text{ or } &
\sqrt{x} & = -2 \text{ (Reject, since } \sqrt{x} \ge 0) \\
x & = 1.8^2 \\
x & = 3.24
\end{align}
\begin{align}
ax^2 + bx + c & = a \left(x^2 + {b \over a}x\right) + c \\
& = a \left[ x^2 + {b \over a}x + \left(b \over 2a\right)^2 - \left(b \over 2a\right)^2 \right] + c \\
& = a \left[ \left(x + {b \over 2a}\right)^2 - {b^2 \over 4a^2} \right] + c \\
& = a \left(x + {b \over 2a}\right)^2 - {ab^2 \over 4a^2} + c \\
& = a \left(x + {b \over 2a}\right)^2 - {b^2 \over 4a} + c \\
\\
\text{Comparing } & \text{with } a(x - h)^2 + k,
\end{align}
\begin{align}
-h & = {b \over 2a} &&& k & = -{b^2 \over 4a} + c \\
h & = -{b \over 2a} \phantom{0} \text{ (Shown)}
&&&
k & = {-b^2 \over 4a} + {c \over 1} \\
&&&& k & = {-b^2 \over 4a} + {4ac \over 4a} \\
&&&& k & = {-b^2 + 4ac \over 4a} \\
&&&& k & = -{b^2 - 4ac \over 4a} \phantom{0} \text{ (Shown)}
\end{align}
\begin{align} 3x^2 - bx + 16 & = 3 \left(x^2 - {b \over 3}x \right) + 16 \\ & = 3 \left[ x^2 - {b \over 3}x + \left(b \over 6\right)^2 - \left(b \over 6\right)^2 \right] + 16 \\ & = 3 \left[ \left(x - {b \over 6}\right)^2 - {b^2 \over 36} \right] + 16 \\ & = 3 \left(x - {b \over 6}\right)^2 - {3b^2 \over 36} + 16 \\ & = 3 \left(x - {b \over 6}\right)^2 - {b^2 \over 12} + 16 \\ \\ \text{For all real} & \text{ values of } x, \\ 3\left(x - {b \over 6}\right)^2 & \ge 0 \\ 3\left(x - {b \over 6}\right)^2 - {b^2 \over 12} + 16 & \ge -{b^2 \over 12} + 16 \\ \\ \text{For } 3x^2 - bx + 16 = 0 & \text{ to have no solutions,} \\ -{b^2 \over 12} + 16 & > 0 \\ -{b^2 \over 12} & > - 16 \\ {b^2 \over 12} & < 16 \\ b^2 & < 12(16) \\ b^2 & < 192 \\ \\ \text{Possible values} & \text{ of } b: 8, -12 \end{align}
\begin{align} & \text{For } y = 8, &&& & \text{For } y = -12, \\ y & = 3 \left(x - {8 \over 6}\right)^2 - {(8)^2 \over 12} + 16 &&& y & = 3 \left(x - {-12 \over 6}\right)^2 - {(-12)^2 \over 12} + 16 \\ y & = 3 \left(x - {4 \over 3}\right)^2 + 10{2 \over 3} &&& y & = 3 (x + 2)^2 + 4 \\ \\ \text{Turning} & \text{ point: } \left({4 \over 3}, 10{2 \over 3}\right) &&& \text{Turning} & \text{ point: } (-2, 4) \\ \\ & \text{Let } x = 0, &&& & \text{Let } x = 0, \\ y & = 3 \left(0 - {4 \over 3}\right)^2 + 10{2 \over 3} &&& y & = 3 (0 + 2)^2 + 4 \\ y & = 16 &&& y & = 16 \end{align}