DAV Class 8 Maths Chapter 7 Practice Worksheet
Algebraic Identities Practice Worksheet
Answer the following Questions
1. If \(x=-1\), find the value of \(x^2-\dfrac{1}{5}\)
Solution
\[ \begin{aligned} &= x^2-\frac15 \\[6pt] &=(-1)^2-\frac15 \\[6pt] &=1-\frac15\\[6pt] &=\frac{5-1}{5} \\ & =\frac{4}{5} \end{aligned} \]Answer \(\color{red}{\dfrac{4}{5}}\)
2. Find the product of \(9a,\ 4ab\) and \(-2a\).
Solution
\[ \begin{aligned} &= 9a \times 4ab \times (-2a) \\[4pt] &= 36 a^2b \times (-2a) \\[4pt] & =-72a^3b \end{aligned} \]Answer \(\color{red}{-72a^3b}\)
3. If \(x=0\) and \(y=0\), find the value of \(x^2+y^2-10\).
Solution
\[ \begin{aligned} &= x^2+y^2-10 \\[4pt] &= 0 + 0 - 10 \\[4pt] & = -10 \end{aligned} \]Answer \(\color{red}{-10}\)
4. Evaluate the following using suitable identity.
(a) \( 9.4\times10.6\)
Solution
\[ \begin{aligned} \color{magenta} 9.4 & = \color{magenta} (10-0.6) \\[4pt] \color{magenta} 10.6 & = \color{magenta} (10 + 0.6) \\ \\ &= 9.4\times10.6 \\[4pt] &= (10-0.6)(10+0.6 )\\[4pt] \implies & \color{green} (a - b)(a + b) = \color{green} a^2 - b^2 \\[6pt] & = (10)^2 - (0.6)^2 \\[4pt] & = 100 - 0.36 \\[4pt] & = 99.64 \end{aligned} \]Answer \( \color{red}{99.64}\)
(b) \(102\times106\)
Solution
\[ \begin{aligned} \color{magenta} 102 &= \color{magenta} (100 + 2) \\[4pt] \color{magenta} 106 &= \color{magenta} (100 + 6) \\ \\ &= 102\times106 \\[4pt] &= (100 + 2)(100 + 6) \\[4pt] \implies &\ \color{green}{(x + a)(x + b)=x^2 + (a + b)x + ab} \\[6pt] &= (100)^2 + (2 + 6)100 + (2)(6) \\[4pt] &= 10000 + 800 + 12 \\[4pt] &= 10812 \end{aligned} \]Answer \( \color{red}{10812}\)
(c) \((99)^2\)
Solution
\[ \begin{aligned} \color{magenta} (99)^2 &= \color{magenta} (100-1)^2 \\[6pt] \implies &\ \color{green}{(a-b)^2=a^2+b^2-2ab} \\[6pt] &= (100)^2 + (1)^2-2(100)(1) \\[4pt] &= 10000+1-200 \\[4pt] &= 9801 \end{aligned} \]Answer \( \color{red}{9801}\)
(d) \((10.5)^2\)
Solution
\[ \begin{aligned} \color{magenta} (10.5)^2 &= \color{magenta} (10+0.5)^2 \\[6pt] \implies &\ \color{green}{(a+b)^2=a^2+b^2+2ab} \\[6pt] &= (10)^2+(0.5)^2+2(10)(0.5) \\[4pt] &= 100+0.25+10 \\[4pt] &= 110.25 \end{aligned} \]Answer \( \color{red}{110.25}\)
(e) \( 95 \times 93\)
Solution
\[ \begin{aligned} \color{magenta} 95 &= \color{magenta} (90 + 5) \\[6pt] \color{magenta} 93 &= \color{magenta} (90 + 3) \\ \\ &= 95 \times 93 \\[6pt] &= (90 + 5)(90 + 3) \\[6pt] \implies &\ \color{green}{(x + a)(x + b)=x^2 + (a + b)x + ab} \\[6pt] &= (90)^2 + (5 + 3)90 + (5)(3) \\[6pt] &= 8100 + 720 + 15 \\[6pt] &= 8820 + 15 \\[6pt] &= 8835 \end{aligned} \]Answer \( \color{red}{8835}\)
(f) \((81)^2-(19)^2\)
Solution
\[ \begin{aligned} \implies &\ \color{green}{a^2-b^2=(a-b)(a+b)} \\[6pt] &= (81)^2-(19)^2 \\[6pt] &= (81-19)(81+19) \\[6pt] &= 62\times100 \\[6pt] &= 6200 \end{aligned} \]Answer \( \color{red}{6200}\)
(g) \(97\times102\)
Solution
\[ \begin{aligned} \color{magenta} 97 &= \color{magenta} (100-3) \\[6pt] \color{magenta} 102 &= \color{magenta} (100+2) \\[6pt] &= (100-3)(100+2) \\[4pt] \implies &\ \color{green}{(x+a)(x+b)=x^2+(a+b)x+ab} \\[6pt] &= (100)^2 + (-3+2)100 + (-3)(2) \\[6pt] &= 10000 + (-1)100 + (-6) \\[6pt] &= 10000 - 100 - 6 \\[6pt] &= 9894 \end{aligned} \]Answer \( \color{red}{9894}\)
(h) \(203\times198\)
Solution
\[ \begin{aligned} \color{magenta} 203 &= \color{magenta} (200+3) \\[6pt] \color{magenta} 198 &= \color{magenta} (200-2) \\[6pt] &= (200+3)(200-2) \\[4pt] \implies &\ \color{green}{(x+a)(x+b)=x^2+(a+b)x+ab} \\[6pt] &= (200)^2 + (3-2)200 + (3)(-2) \\[6pt] &= 40000 + (1)200 + (-6) \\[6pt] &= 40000 + 200 - 6 \\[6pt] &= 40194 \end{aligned} \]Answer \( \color{red}{40194}\)
(i) \(104\times92\)
Solution
\[ \begin{aligned} \color{magenta} 104 &= \color{magenta} (100+4) \\[6pt] \color{magenta} 92 &= \color{magenta} (100-8) \\[6pt] &= (100+4)(100-8) \\[4pt] \implies &\ \color{green}{(x+a)(x+b)=x^2+(a+b)x+ab} \\[6pt] &= (100)^2 + (4-8)100 + (4)(-8) \\[4pt] &= 10000 + (-4)100 + (-32) \\[4pt] &= 10000 - 400 - 32 \\[4pt] &= 9568 \end{aligned} \]Answer \( \color{red}{9568}\)
5. Evaluate the following using suitable identity.
(a) \((8x+3y)^2\)
Answer \(\color{red}{64x^2+48xy+9y^2}\)
(b) \((2m^2+3n^2)^2\)
Solution
\[ \begin{aligned} \implies&\ \color{green}{(a+b)^2=a^2+2ab+b^2} \\[6pt] & = (2m^2+3n^2)^2 \\[6pt] &=(2m^2)^2+2(2m^2)(3n^2)+(3n^2)^2 \\[6pt] &=4m^4+12m^2n^2+9n^4 \end{aligned} \]Answer \(\color{red}{4m^4+12m^2n^2+9n^4}\)
(c) \((5a-4b)^2\)
Solution
\[ \begin{aligned} \implies&\ \color{green}{(a-b)^2=a^2-2ab+b^2} \\[6pt] & = (5a-4b)^2 \\[6pt] &=(5a)^2-2(5a)(4b)+(4b)^2 \\[6pt] &=25a^2-40ab+16b^2 \end{aligned} \]Answer \(\color{red}{25a^2-40ab+16b^2}\)
(d) \((6mn-5pq)^2\)
Solution
\[ \begin{aligned} \implies&\ \color{green}{(a-b)^2=a^2-2ab+b^2} \\[6pt] & = (6mn-5pq)^2 \\[6pt] &=(6mn)^2-2(6mn)(5pq)+(5pq)^2 \\[6pt] &=36m^2n^2-60mnpq+25p^2q^2 \end{aligned} \]Answer \(\color{red}{36m^2n^2-60mnpq+25p^2q^2}\)
(e) \((4p^2+q^2)(4p^2-q^2)\)
Solution
\[ \begin{aligned} \implies&\ \color{green}{(a+b)(a-b)=a^2-b^2} \\[6pt] & = (4p^2+q^2)(4p^2-q^2) \\[6pt] &=(4p^2)^2-(q^2)^2 \\[6pt] &=16p^4-q^4 \end{aligned} \]Answer \(\color{red}{16p^4-q^4}\)
(f) \((a^3+b^3)(a^3-b^3)\)
Solution
\[ \begin{aligned} \implies&\ \color{green}{(a+b)(a-b)=a^2-b^2} \\[6pt] &= (a^3+b^3)(a^3-b^3) \\[6pt] &=(a^3)^2-(b^3)^2 \\[6pt] &=a^6-b^6 \end{aligned} \]Answer \(\color{red}{a^6-b^6}\)
6. Find the product using suitable identity.
(a) \((x + 5)(x + 11)\)
Solution
\[ \begin{aligned} \color{green}(x + a)(x + b) & = \color{green} x^2 + (a + b)x + ab \\[6pt] (x + 5)(x + 11) &= x^2 + (5 + 11)x + (5)(11) \\[6pt] &= x^2 + 16x + 55 \end{aligned} \]Answer \(\color{red} x^2 + 16x + 55 \)
(b) \( (a - 15)(a + 7)\)
Solution
\[ \begin{aligned} \color{green}(x + a)(x + b) & = \color{green} x^2 + (a + b)x + ab \\[6pt] (a - 15)(a + 7) &= a^2 + (-15 + 7)a + (-15)(7) \\[6pt] &= a^2 + (-8)a + (-105) \\[6pt] &= a^2 - 8a -105 \end{aligned} \]Answer \(\color{red} a^2 - 8a -105 \)
(c) \((p - 9)(p + 8)\)
Solution
\[ \begin{aligned} \color{green}(x + a)(x + b) & = \color{green} x^2 + (a + b)x + ab \\[6pt] (p - 9)(p + 8) &= p^2 + (-9 + 8)p + (-9)(8) \\[6pt] &= p^2 + (-1)p + (-72) \\[6pt] &= p^2 - p - 72 \end{aligned} \]Answer \(\color{red}{p^2 - p - 72}\)
(d) \((m - 4)(m - 14)\)
Solution
\[ \begin{aligned} \color{green}(x + a)(x + b) & = \color{green} x^2 + (a + b)x + ab \\[6pt] (m - 4)(m - 14) &= m^2 + (-4 - 14)m + (-4)(-14) \\[6pt] &= m^2 - 18m + 56 \end{aligned} \]Answer \(\color{red}{m^2 - 18m + 56}\)
(e) \((y - 6)(y - 8)\)
Solution
\[ \begin{aligned} \color{green}(x + a)(x + b) & = \color{green} x^2 + (a + b)x + ab \\[6pt] (y - 6)(y - 8) &= y^2 + (-6 - 8)y + (-6)(-8) \\[6pt] &= y^2 - 14y + 48 \end{aligned} \]Answer \(\color{red}{y^2 - 14y + 48}\)
(f) \((z + 6)(z - 5)\)
Solution
\[ \begin{aligned} \color{green}(x + a)(x + b) & = \color{green} x^2 + (a + b)x + ab \\[6pt] (z + 6)(z - 5) &= z^2 + (6 - 5)z + (6)(-5) \\[6pt] &= z^2 + z - 30 \end{aligned} \]Answer \(\color{red}{z^2 + z - 30}\)
(g) \((x + 12)(x - 7)\)
Solution
\[ \begin{aligned} \color{green}(x + a)(x + b) & = \color{green} x^2 + (a + b)x + ab \\[6pt] (x + 12)(x - 7) &= x^2 + (12 - 7)x + (12)(-7) \\[6pt] &= x^2 + 5x - 84 \end{aligned} \]Answer \(\color{red}{x^2 + 5x - 84}\)
(h) \((p + 18)(p - 3)\)
Solution
\[ \begin{aligned} \color{green}(x + a)(x + b) & = \color{green} x^2 + (a + b)x + ab \\[6pt] (p + 18)(p - 3) &= p^2 + (18 - 3)p + (18)(-3) \\[6pt] &= p^2 + 15p - 54 \end{aligned} \]Answer \(\color{red}{p^2 + 15p - 54}\)
(i) \((a + 13)(a - 9)\)
Solution
\[ \begin{aligned} \color{green}(x + a)(x + b) & = \color{green} x^2 + (a + b)x + ab \\[6pt] (a + 13)(a - 9) &= a^2 + (13 - 9)a + (13)(-9) \\[6pt] &= a^2 + 4a - 117 \end{aligned} \]Answer \(\color{red}{a^2 + 4a - 117}\)
7. Factorise the following.
(a) \(m^2-2m-15\)
Solution
\[ \begin{aligned} &= m^2\ {\color{magenta}-\,2m}-15 \\[4pt] &= m^2\ {\color{magenta}-\,5m+3m}-15 \\[4pt] &= m(m-5)+3(m-5) \\[4pt] &= \color{red}{(m-5)(m+3)} \end{aligned} \](b) \(x^2-1-2y-y^2\)
Solution
\[ \begin{aligned} &= x^2-1-2y-y^2 \\[4pt] &= x^2-y^2 -2y -1 \\[4pt] &= x^2-\big(y^2+2y+1\big) \\[4pt] &= (x)^2-(y+1)^2 \\[4pt] &= [x-(y+1)][x+(y+1)] \\[4pt] &= \color{red}{(x-y-1)(x+y+1)} \end{aligned} \](c) \(121p^2+16q^2-88pq\)
Solution
\[ \begin{aligned} &= 121p^2+16q^2-88pq \\[4pt] &= (11p)^2+(4q)^2-2(11p)(4q) \\[4pt] &= (11p-4q)^2 \\[4pt] &= \color{red}{(11p-4q)(11p-4q)} \end{aligned} \](d) \(64a^2+49b^2+112ab\)
Solution
\[ \begin{aligned} &= 64a^2+49b^2+112ab \\[4pt] &= (8a)^2+(7b)^2+2(8a)(7b) \\[4pt] &= (8a+7b)^2 \\[4pt] &= \color{red}{(8a+7b)(8a+7b)} \end{aligned} \](e) \(y^2-25\)
Solution
\[ \begin{aligned} &= (y)^2-(5)^2 \\[4pt] &= \color{red}{(y-5)(y+5)} \end{aligned} \](f) \(1-6z+9z^2\)
Solution
\[ \begin{aligned} &= 9z^2 - 6z + 1 \\[4pt] &= (3z)^2-2(3z)(1)+(1)^2 \\[4pt] & = (3z-1)^2 \\[4pt] &= \color{red}{(3z-1)(3z-1)} \end{aligned} \](g) \(x^2+2x-63\)
Solution
\[ \begin{aligned} &= x^2\ {\color{magenta}+\,2x}-63 \\[4pt] &= x^2\ {\color{magenta}+\,9x-7x}-63 \\[4pt] &= x(x+9)-7(x+9) \\[4pt] &= \color{red}{(x+9)(x-7)} \end{aligned} \](h) \(x^2-x-12\)
Solution
\[ \begin{aligned} &= x^2\ {\color{magenta}-\,x}-12 \\[4pt] &= x^2\ {\color{magenta}-\,4x+3x}-12 \\[4pt] &= x(x-4)+3(x-4) \\[4pt] &= \color{red}{(x-4)(x+3)} \end{aligned} \](i) \(p^2-17p-38\)
Solution
\[ \begin{aligned} &= p^2\ {\color{magenta}-\,17p}-38 \\[4pt] &= p^2\ {\color{magenta}-\,19p+2p}- 38 \\[4pt] &= p(p-19)+2(p-19) \\[4pt] &= \color{red}{(p-19)(p+2)} \end{aligned} \](j) \(4x^2+12x+9\)
Solution
\[ \begin{aligned} &= 4x^2+12x+9 \\[4pt] &= (2x)^2+2(2x)(3)+(3)^2 \\[4pt] & = (2x+3)^2 \\[4pt] &= \color{red}{(2x+3)(2x+3)} \end{aligned} \](k) \(y^2+3y+y+3\)
Solution
\[ \begin{aligned} &= y^2+3y+y+3 \\[4pt] &= y(y+3)+1(y+3) \\[4pt] &= \color{red}{(y+1)(y+3)} \end{aligned} \](l) \(x^2+y-xy-x\)
Solution
\[ \begin{aligned} &= x^2+y-xy-x \\[4pt] &= x^2-x-xy+y \\[4pt] &= x(x-1)-y(x-1) \\[4pt] &= \color{red}{(x-1)(x-y)} \end{aligned} \]8. If \(x^2+y^2=57\) and \(xy=16\), then find the value of \((x-y)^2\).
Solution
\[ \begin{aligned} (x-y)^2 & = x^2+y^2-2xy \\[6pt] (x-y)^2 &= 57 - 2(16) \\[4pt] (x-y)^2 &= 57-32 \\[4pt] (x-y)^2 &= 25 \end{aligned} \]Answer \((x-y)^2=\;\color{red}{25}\)
9. If \(\left(a+\dfrac{1}{a}\right)=\dfrac{17}{4}\), then find \(\left(a-\dfrac{1}{a}\right)\).
Solution
\[ \begin{aligned} \left(a+\frac{1}{a}\right) &= \frac{17}{4} \\[6pt] \text{Square on } & \text{both sides} \\[6pt] \left(a+\frac{1}{a}\right)^2 &= \left(\frac{17}{4}\right)^2 \\[6pt] a^2+ \frac{1}{a^2} + 2 \times \cancel a \times \frac{1}{\cancel a} &= \frac{289}{16} \\[6pt] a^2+ \frac{1}{a^2} + 2 &= \frac{289}{16} \\[6pt] a^2+ \frac{1}{a^2} &= \frac{289}{16} - 2 \\[6pt] a^2+ \frac{1}{a^2} &= \frac{289 - 32}{16} \\[6pt] \color{green} a^2+ \frac{1}{a^2} &= \color{green} \frac{257}{16} \\[8pt] \left(a-\frac{1}{a}\right)^2 &= a^2+ \frac{1}{a^2} - 2 \times \cancel a \times \frac{1}{\cancel a} \\[6pt] \left(a-\frac{1}{a}\right)^2 &= a^2+ \frac{1}{a^2} - 2 \\[6pt] \left(a-\frac{1}{a}\right)^2 &= \frac{257}{16} - 2 \\[6pt] \left(a-\frac{1}{a}\right)^2 &= \frac{257 - 32}{16} \\[6pt] \left(a-\frac{1}{a}\right)^2 &= \frac{225}{16} \\[6pt] \left(a-\frac{1}{a}\right)^2 &= \left(\frac{15}{4}\right)^2 \\[6pt] a-\frac{1}{a} &= \frac{15}{4} \\[6pt] \end{aligned} \]Answer \(\left(a-\dfrac{1}{a}\right)=\;\color{red}{\dfrac{15}{4}}\)
10. The sum of \((x+3)\) observations is \((x^4-81)\). Find the mean of the observations.
Solution
\[ \begin{aligned} \text{No. of observations} &= x+3 \\[4pt] \text{Sum of observations} &= x^4-81 \\[6pt] \color{green}\text{Mean} &= \color{green}\frac{\text{Sum of observations}}{\text{No. of observations}} \\[6pt] &= \frac{x^4-81}{x+3} \\[6pt] &= \frac{(x^2)^2-(9)^2}{(x+3)} \\[6pt] &= \frac{(x^2 - 9)(x^2 + 9)}{(x+3)} \\[6pt] &= \frac{[(x)^2 - (3)^2](x^2 + 9)}{(x+3)} \\[6pt] &= \frac{(x -3)\cancel{(x+3)}(x^2 + 9)}{\cancel{(x+3)}} \\[6pt] \text{Mean} &= (x-3)(x^2+9) \end{aligned} \]Answer Mean \(=\;\color{red}{(x-3)(x^2+9)}\)
11. The area of the circle is \(\pi x^2+10\pi x+25\pi\). Find the radius of the circle.
Solution
\[ \begin{aligned} \color{green} \text{Area of a circle} &= \pi x^2+10\pi x+25\pi \\[6pt] \color{green}\pi r^2 &= \pi\,(x^2+10x+25) \\[6pt] \pi r^2 &= \pi\,[(x)^2+ 2(x)(5)+(5)^2] \\[6pt] \pi r^2 &= \pi\,(x+5)^2 \\[6pt] r^2 &= \frac{\cancel{\pi}(x+5)^2}{\cancel{\pi}}\\[6pt] r^2 &= (x+5)^2 \\[6pt] r &= \sqrt{(x+5)^2} \\[6pt] r &= x+5 \end{aligned} \]Answer Radius \(=\;\color{red}{x+5}\)
12. If \(6x-7y=18\) and \(xy=-3\), find the value of \(36x^2+49y^2\).
Solution
\[ \begin{aligned} 6x-7y &= 18 \\[6pt] \text{Square on } & \text{both sides} \\[6pt] (6x-7y)^2 &= (18)^2 \\[4pt] (6x)^2+(7y)^2 - 2(6x)(7y) &= 324 \\[4pt] 36x^2+49y^2 -84xy &= 324 \\[4pt] 36x^2+49y^2 -84(-3) &= 324 \\[4pt] 36x^2+49y^2 + 252 &= 324 \\[4pt] 36x^2+49y^2 &= 324 - 252 \\[4pt] 36x^2+49y^2 &= 72 \end{aligned} \]Answer \(36x^2+49y^2=\;\color{red}{72}\)
13. If \(5a-2b=7\) and \(ab=2\), then find the value of \((5a+2b)^2\).
Solution
\[ \begin{aligned} 5a-2b &= 7 \\[6pt] \text{Square on } & \text{both sides} \\[6pt] (5a-2b)^2 &= 7^2 \\[6pt] 25a^2+4b^2-20ab &= 49 \\[6pt] 25a^2+4b^2-20(2) &= 49 \\[6pt] 25a^2+4b^2-40 &= 49 \\[6pt] 25a^2+4b^2 &= 49+40 \\[6pt] \color{green} 25a^2+4b^2 &=\color{green} 89 \\[8pt] (5a+2b)^2 &= 25a^2+4b^2+20ab \\[6pt] (5a+2b)^2 &= 89+20(2) \\[6pt] (5a+2b)^2 &= 129 \end{aligned} \]Answer \((5a+2b)^2=\;\color{red}{129}\)
14. If \(m^2+n^2=9\) and \(mn=4\), then find the value of \(3(m+n)^2\).
Solution
\[ \begin{aligned} & = 3(m+n)^2 \\[4pt] &= 3[m^2+n^2+2mn] \\[4pt] &= 3[9+2(4)] \\[4pt] &= 3[9+8] \\[4pt] &= 3 \times 17 \\[4pt] &= 51 \end{aligned} \]Answer \(3(m+n)^2=\;\color{red}{51}\)
15. Simplify using suitable identity: \((5xy+4y)^2-(5xy-4y)^2\).
Solution
\[ \begin{aligned} & \color{magenta}(a)^2-(b)^2 = (a+b)(a-b) \\[6pt] & = (5xy+4y)^2-(5xy-4y)^2 \\[6pt] & = [5xy+4y +(5xy-4y)][5xy+4y -(5xy-4y)] \\[6pt] & = (5xy + \cancel{4y} +5xy- \cancel{4y})( \cancel{5xy}+4y -\cancel{5xy} +4y) \\[6pt] & = (5xy +5xy )( 4y +4y) \\[6pt] & = (10xy) (8y) \\[6pt] &= 80xy^2 \end{aligned} \]Answer \(\color{red}{80xy^2}\)
16. If \((4p-3q)=9\) and \(pq=3\), then find the value of \((4p+3q)^2\).
Solution
\[ \begin{aligned} 4p-3q &= 9 \\[4pt] \text{Square on } & \text{both sides} \\[6pt] (4p-3q)^2 &= 9^2 \\[4pt] 16p^2+9q^2-24pq &= 81 \\[4pt] 16p^2+9q^2-24(3) &= 81 \\[4pt] 16p^2+9q^2-72 &= 81 \\[4pt] 16p^2+9q^2 &= 81+72 \\[4pt] \color{green} 16p^2+9q^2 & = \color{green} 153 \\ \\ (4p+3q)^2 &= 16p^2+9q^2+24pq \\[4pt] (4p+3q)^2 &= 153+24(3) \\[4pt] (4p+3q)^2 &= 153+72 \\[4pt] (4p+3q)^2 &= 225 \\[4pt] \end{aligned} \]Answer \((4p+3q)^2=\;\color{red}{225}\)
17. If \(5x-4y=12\) and \(xy=2\) then find the value of: (i) \(25x^2+16y^2\) (ii) \((5x+4y)^2\)
Solution
\[ \begin{aligned} \textbf{(i)} \ \color{blue} 25x^2+16y^2 \\[6pt] 5x-4y &= 12 \\[6pt] \text{Square on } & \text{both sides} \\[6pt] (5x-4y)^2 &= 12^2 \\[6pt] 25x^2+16y^2-40xy &= 144 \\[6pt] 25x^2+16y^2-40(2) &= 144 \\[6pt] 25x^2+16y^2- 80 &= 144 \\[6pt] 25x^2+16y^2 &= 144 + 80 \\[6pt] \color{green}{25x^2+16y^2} &= \color{green}{224} \\ \\ \textbf{(ii)} \ \color{blue} (5x+4y)^2 \\[6pt](5x+4y)^2 &= 25x^2+16y^2+40xy \\[6pt] (5x+4y)^2 &= 224+40(2) \\[4pt] (5x+4y)^2 &= 224+80\\[4pt] \color{green}(5x+4y)^2 &= \color{green} 304 \end{aligned} \]Answer (i) \( \color{red}{224} \ \) (ii) \(\color{red}{304}\)
18. If \(xy^2k=(3xy+4y)^2-(3xy-4y)^2\), then find the value of \(k\).
Solution
\[ \begin{aligned} & \big[ {\color{magenta}{a^2-b^2=(a-b)(a+b)}}\big] \\[8pt] xy^2k & = (3xy+4y)^2-(3xy-4y)^2 \\[4pt] xy^2k &= \big[3xy+4y-( 3xy-4y) \big] \big[3xy+4y + (3xy-4y )\big] \\[4pt] xy^2k &= \big(\cancel{3xy} + 4y - \cancel{3xy} + 4y \big) \big(3xy + \cancel{4y} + 3xy- \cancel{4y}\big) \\[4pt] xy^2k &= (8y)(6xy) \\[4pt] xy^2k &= 48xy^2 \\[6pt] k &= \frac{48 \cancel{xy^2}}{\cancel{xy^2}} \\[6pt] k &= 48 \\[4pt] \end{aligned} \]Answer \(k=\;\color{red}{48}\)
19. If \(x^2+\dfrac{1}{x^2}=6\), then find the value of \(x^4+\dfrac{1}{x^4}\).
Solution
\[ \begin{aligned} x^2+\dfrac{1}{x^2} & = 6 \\[6pt] \text{Square on } & \text{both sides} \\[6pt] \left(x^2+\frac{1}{x^2}\right)^2 &= 6^2 \\[6pt] x^4+\frac{1}{x^4}+2\times\cancel{x^2}\times\frac{1}{\cancel{x^2}} &= 36 \\[6pt] x^4+\frac{1}{x^4}+2 &= 36 \\[6pt] x^4+\frac{1}{x^4} &= 36-2 \\[6pt] x^4+\frac{1}{x^4} &= 34 \end{aligned} \]Answer \(x^4+\dfrac{1}{x^4}=\;\color{red}{34}\)
20. If \(\left(x+\dfrac{1}{x}\right)=8\), then find the value of \(x^2+\dfrac{1}{x^2}\).
Solution
\[ \begin{aligned} \left(x+\dfrac{1}{x}\right) & = 8 \\[6pt] \text{Square on } & \text{both sides} \\[6pt] \left(x+\frac{1}{x}\right)^2 &= 8^2 \\[6pt] x^2+\frac{1}{x^2}+2\times\cancel{x}\times\frac{1}{\cancel{x}} &= 64 \\[6pt] x^2+\frac{1}{x^2}+2 &= 64 \\[6pt] x^2+\frac{1}{x^2} &= 64-2 \\[6pt] x^2+\frac{1}{x^2} &= 62 \end{aligned} \]Answer \(x^2+\dfrac{1}{x^2}=\;\color{red}{62}\)
21. If \(\left(a-\dfrac{1}{a}\right)=5\), then find the value of \(a^2+\dfrac{1}{a^2}\).
Solution
\[ \begin{aligned} \left(a-\dfrac{1}{a}\right)&=5 \\[6pt] \text{Square on } & \text{both sides} \\[6pt] \left(a-\frac{1}{a}\right)^2 &= 5^2 \\[6pt] a^2+\frac{1}{a^2}-2\times\cancel{a}\times\frac{1}{\cancel{a}} &= 25 \\[6pt] a^2+\frac{1}{a^2}-2 &= 25 \\[6pt] a^2+\frac{1}{a^2} &= 25+2 \\[6pt] a^2+\frac{1}{a^2} &= 27 \end{aligned} \]Answer \(a^2+\dfrac{1}{a^2}=\;\color{red}{27}\)