S3 E Maths Textbook Solutions >> New Discovering Mathematics 3A Chapter 1 & 2 Solutions >>
Ex 1.1
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Solutions
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(a)
x2−4x+3=0(x−1)(x−3)=0 x−1=0 or x−3=0x=1x=3
(b)
x2−2x−8=0(x+2)(x−4)=0 x+2=0 or x−4=0x=−2x=4
(c)
2x2−3x+1=0(2x−1)(x−1)=0 2x−1=0 or x−1=02x=1x=1x=12
(d)
15x2+8x+1=0(5x+1)(3x+1)=0 5x+1=0 or 3x+1=05x=−13x=−1x=−15x=−13
(e)
2x2+x−6=0(2x−3)(x+2)=0 2x−3=0 or x+2=02x=3x=−2x=32
(f)
3x2−2x−5=0(3x−5)(x+1)=0 3x−5=0 or x+1=03x=5x=−1x=53
(g)
4x2−12x+9=0(2x−3)(2x−3)=0(2x−3)2=02x−3=02x=3x=32
(h)
25x2−16=0(5x)2−(4)2=0(5x+4)(5x−4)=0000000[a2−b2=(a+b)(a−b)] 5x+4=0 or 5x−4=05x=−45x=4x=−45x=45
(i)
14x2−29x=1514x2−29x−15=0(2x−5)(7x+3)=0 2x−5=0 or 7x+3=02x=57x=−3x=52x=−37
(j)
x2−2x=5x−12x2−2x−5x+12=0x2−7x+12=0(x−3)(x−4)=0 x−3=0 or x−4=0x=3x=4
(a)
k=(22)2=12=1
(b)
k=(42)2=22=4
(c)
k=(82)2=42=16
(d)
k=(72)2=494
(e)
k=(12)2=14
(f)
k=(112)2=1214
(a)
x2+2x+5=x2+2x+(22)2−(22)2+5=x2+2x+12−12+5=(x+1)2−1+5=(x+1)2+4
(b)
x2−6x−3=x2−6x+(62)2−(62)2−3=x2−6x+32−32−3=(x−3)2−9−3=(x−3)2−12
(c)
x2+5x−6=x2+5x+(52)2−(52)2−6=(x+52)2−254−6=(x+52)2−494
(d)
x2−9x+4=x2−9x+(92)2−(92)2+4=(x−92)2−814+4=(x−92)2−654
(a)
(x−1)2=25x−1=±√25x−1=±5 x−1=5 or x−1=−5x=5+1x=−5+1x=6x=−4
(b)
(x+3)2=49x+3=±√49x+3=±7 x+3=7 or x+3=−7x=7−3x=−7−3x=4x=−10
(c)
(x+12)2=1x+12=±√1x+12=±1 x+12=1 or x+12=−1x=1−12x=−1−12x=12x=−32
(d)
(x−52)2=94x−52=±√94x−52=±32 x−52=32 or x−52=−32x=32+52x=−32+52x=4x=1
(e)
(x+4)2=7x+4=±√7 x+4=√7 or x+4=−√7x=√7−4x=−√7−4x≈−1.35x≈−6.65
(f)
(x−3)2=10x−3=±√10 x−3=√10 or x−3=−√10x=√10+3x=−√10+3x≈6.16x≈−0.162
(a)
x2+8x+12=x2+8x+(82)2−(82)2+12=x2+8x+42−42+12=(x+4)2−16+12=(x+4)2−4
(b)
x2+8x+12=0(x+4)2−4⏟From (a)=0(x+4)2=4x+4=±√4x+4=±2 x+4=2 or x+4=−2x=2−4x=−2−4x=−2x=−6
(a)
x2−6x+4=x2−6x+(62)2−(62)2+4=x2−6x+32−32+4=(x−3)2−9+4=(x−3)2−5
(b)
x2−6x+4=0(x−3)2−5⏟From (a)=0(x−3)2=5x−3=±√5 x−3=√5 or x−3=−√5x=√5+3x=−√5+3x≈5.24x≈0.764
(a)
x2+2x+3=x2+2x+(22)2−(22)2+3=x2+2x+12−12+3=(x+1)2−1+3=(x+1)2+2
(b)
x2+2x+3=0(x+1)2+2⏟From (a)=0(x+1)2=−2x+1=±√−2No real roots for √−2∴
(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}