ML aggarwal class 8 chapter 2 Exponents and powers Exercise 2.1

Exercise 2.1

Question 1

Sol :
(i)
 $\begin{aligned}\left(\frac{3}{5}\right)^{2} &=\left(\frac{5}{3}\right)^{2} \\ &=\frac{5 \times 5}{3 \times 3} \\ &=\frac{25}{9} \end{aligned}$  $\left(\because(p / q)^{-m}=\left(\frac{q}{p}\right)^{m}\right)$

(ii) 
$\begin{aligned}(-3)^{-3} &=\frac{1}{(-3)^{3}} \quad\left(\because(a)^{-m}=\frac{1}{(a)^{m}}\right) \\ &=\frac{1}{(-3) \times(-3) \times(-3)} \\ &=\frac{1}{-27}=\frac{-1}{27} \end{aligned}$

(iii)
 $\begin{aligned}\left(\frac{2}{7}\right)^{-4} &=\left(\frac{7}{2}\right)^{4} \\ &=\frac{7 \times 7 \times 7 \times 7}{2 \times 2 \times 2 \times 2} \\ &=\frac{2401}{16} \end{aligned}$

Question 2

Sol :
(i)
$\begin{aligned}\left[(2)^{-1}+(4)^{-1}+(3)^{-1}\right]^{-1} &=\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{3}\right]^{-1} \quad\left(\because(a)^{-m}=\frac{1}{a^{m}}\right) \\ &=\left[\frac{6+3+4}{12}\right]^{-1} \\ &=\left(\frac{13}{12}\right)^{-1} \\ &=\frac{12}{13} \end{aligned}$

(ii)
$\begin{aligned}\left[(4)^{-1}-(5)^{-1}\right]^{2} \times\left(\frac{5}{8}\right)^{-1} &=\left(\frac{1}{4}-\frac{1}{5}\right)^{2} \times\left(\frac{8}{5}\right)^{1} \\ &=\left(\frac{5-4}{20}\right)^{2} \times \frac{8}{5} \\ &=\frac{1}{20 \times 20} \times \frac{8}{5} \\ &=\frac{1}{250} \end{aligned}$

(iii)
 $\begin{aligned}\left[4^{0}+4^{2}-2^{3}\right] \times 3^{-2} &=[1+16-8] \times \frac{1}{3^{2}} \\ &=(9) \times \frac{1}{9} . \\ &=1 . \end{aligned}$

(iv)
 $\begin{aligned}\left.[5)^{2}-\left(\frac{1}{4}\right)^{-2}\right] \times\left(\frac{3}{4}\right)^{-2} &=\left[(25)-(4)^{2}\right] \times\left(\frac{4}{3}\right)^{2} \\ &=[25-16] \times \frac{16}{9} \\ &=9 \times \frac{16}{9} \\ &=16 . \end{aligned}$

Question 3 

Sol :
(i)  As the multiplicative inverse of $a^{m}=a^{m}$,

$\therefore$ the multiplicative inverse of $\left(\frac{81}{16}\right)^{-\frac{3}{4}}=\left(\frac{81}{16}\right)^{\frac{3}{4}}$

$\begin{aligned}\left(\frac{81}{16}\right)^{\frac{3}{4}} &=\left(\frac{3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2}\right)^{\frac{3}{4}} \\ &=\left(\frac{3^{4}}{2^{4}}\right)^{\frac{3}{4}} \\ &=\left(\frac{3}{2}\right)^{4 \times \frac{3}{4}} \\ &=\left(\frac{3}{2}\right)^{3}=\frac{3 \times 3 \times 3}{2 \times 2 \times 2} \\ &=\frac{27}{8}  \end{aligned}$

(ii) $\begin{aligned}\left\{\left(\frac{-3}{2}\right)^{-4}\right]_{}^{1 / 2} &=\left\{\left(\frac{-3}{2}\right)^{-4 \times \frac{1}{2}}\right\}\\ &=\left(\frac{-3}{2}\right)^{-2} \end{aligned}$

Multiplicative invorse of $\left(\frac{-3}{2}\right)^{-2}=(-3 / 2)^{2}$

$\left(\frac{-3}{2}\right)^{2}=\frac{-3 \times -3}{2 \times 2}=\frac{9}{4}$

(iii) $\left(\frac{5}{7}\right)^{-2} \times\left(\frac{5}{7}\right)^{4} \div\left(\frac{5}{7}\right)^{3}=\left(\frac{5}{7}\right)^{-2+4-3}\left(\begin{array}{c}-a^{m} \times a^{n}=a^{m+n} \\ a^{m} \div a^{n}=a^{m-n}\end{array}\right)$
$=\left(\frac{5}{7}\right)^{-1}$
Multiplicative invarse of $\left(\frac{5}{7}\right)^{-1}=\left(\frac{5}{7}\right) $

Question 4

Sol :
(i) 
$\begin{array}{rlr}(16)^{-2} & =(2 \times 2 \times 2 \times 2)^{-2} & \\ & =\left(2^{4}\right)^{-2} & \\ & =2^{4 x-2} & \left(\because\left(a^{m}\right)^{n}=a^{m n}\right) \\ & =2^{-8} .\end{array}$

(ii)
 $\begin{aligned}(125)^{-4} &=(5 \times 5 \times 5)^{-4} \\ &=\left((5)^{3}\right)^{-4} \\ &=5^{3 \times(-4)} \\ &=5^{(-12)} \end{aligned}$

Question 5

Sol :
(i) 
$\begin{aligned} 2789.453 &=2 \times 1000+7 \times 100+8 \times 10+9 \times 1+\frac{4}{10}+\frac{5}{100}+\frac{3}{1000} \\ &=2 \times 10^{3}+7 \times 10^{2}+8 \times 10^{1}+9 \times 10^{\circ}+4 \times 10^{-1}+5 \times 10^{-2}+3 \times 10^{-3} \end{aligned}$

(ii) 3007.805
$=3 \times 1000+0 \times 100+0 \times 10+7 \times 1+\frac{8}{10}+\frac{0}{100}+\frac{5}{1000}$
$=3 \times 10^{3}+0 \times 10^{2}+0 \times 10^{1}+7 \times 10^{\circ}+8 \times 10^{-1}+0 \times 10^{-2}+5 \times 10^{-3}$

Question 6

Sol :
(i)
 $\begin{aligned}\left.\left[\left(\frac{5}{7}\right)^{2}\right)^{-1}\right]^{-3} &=\left(\frac{5}{7}\right)^{2 x-1 x-3} \quad\left(\because\left(a^{m}\right)^{n}=a^{m \times n}\right) \\ &=\left(\frac{5}{7}\right)^{6} \end{aligned}$

(ii)
 $\left(\frac{2}{7}\right)^{2} \times\left(\frac{7}{2}\right)^{-3} \div\left\{\left(\frac{7}{5}\right)^{-2}\right\}^{-4}$
$=\left(\frac{2}{7}\right)^{2} \times\left(\frac{2}{7}\right)^{3} \div\left(\frac{7}{5}\right)^{(-2) \times(-4)} \quad\left[\because\left(\frac{p}{q}\right)^{-m}=\left(\frac{q}{p}\right)^{m}\right]$
$=\left(\frac{2}{7}\right)^{5} \div\left(\frac{7}{5}\right)^{8}$
$=\left(\frac{2}{7}\right)^{5} \times\left(\frac{5}{7}\right)^{8}$

$=\frac{2^{5} \times 5^{8}}{7^{5} \times 7^{8}}=\frac{2^{5} \times 5^{8}}{7^{(5+8)}} \quad\left(\because a^{m} \times a^{n}=a^{m+n}\right)$
$=\frac{2^{5} \times 5^{8}}{7^{13}}$

(iii)
 $\begin{aligned}\left(\frac{4}{5}\right)^{2} & \times(5)^{4} \times\left(\frac{2}{5}\right)^{-2} \div\left(\frac{5}{2}\right)^{-3} \\ &=\left(\frac{4}{5}\right)^{2} \times(5)^{4} \times\left(\frac{5}{2}\right)^{2} \div\left(\frac{2}{5}\right)^{3} \\ &=\frac{4^{2} \times 5^{4}}{5^{2}} \times \frac{5^{2}}{2^{2}} \times \frac{5^{3}}{2^{3}} \\ &=\frac{5^{4+2+3-2} \times\left(2^{4}\right)^{2}}{2^{2+3}}=\frac{5^{7}}{2^{5-4}}=\frac{5^{7}}{2^{1}} . \end{aligned}$

(iv) 
$\begin{aligned} \frac{8^{-1} \times 5^{3}}{2^{-4}} &=\frac{\left(2^{3}\right)^{-1} \times 5^{3}}{2^{-4}} \\ &=2^{-3+4} \times 5^{3} \\ &=5^{3} \times 2^{1}  \end{aligned}$

Question 7

Sol :
(i) $\left((-2)^{3}\right)^{2}+(5)^{-3} \div(5)^{-5}-(-1 / 2)^{0}$
$=(-2)^{3 \times 2}+(5)^{-3} \times(5)^{+5}-1 \quad\left(a^{0}=1 ;\left(a^{m}\right)^{n}=a^{m n}\right)$
$=2^{6}+5^{2}-1$
=64+25-1
=88
$=8 \times 11$
$=2^{3} \times 11$

(ii) $3^{-5} \times 3^{2} \div 3^{-6}+\left(2^{2} \times 3^{2}\right)^{2}+\left(\frac{2}{3}\right)^{-1}+2^{-1}+\left(\frac{1}{19}\right)^{-1}$
$=3^{-5+2} \times 3^{6}+2^{4} \times 3^{2}+\left(\frac{3}{2}\right)^{1}+\left(\frac{1}{2}\right)^{\prime}+(19)^{1}$

$=3^{-3+6}+2^{4} \times 3^{2}+\frac{3+1}{2}+19$

$=3^{4}+2^{4} \times 3^{2}+\frac{4}{2}+19$

$=192=3 \times 64$

$=27+144+2+19$

$=192=3 \times 64$

$=2^{6} \times 3$

Question 8

Sol : 

(i) 
$\begin{aligned} 5^{3} \times\left(\frac{4}{5}\right)^{3} &=5^{3} \times \frac{4^{3}}{5^{3}} \\ &=4^{3} \times 5^{3-3} \\ &=4^{3} \\ &=\frac{1}{(4)^{-3}} \end{aligned}$

(ii)
 $\begin{aligned}\left.\left(\frac{3}{7}\right)^{-2}\right]^{-3} &=\left(\frac{3}{7}\right)^{-2 x-3} \quad\left(\because\left(a^{m}\right)^{n}=a^{m \times n}\right) \\ &=\left(\frac{3}{7}\right)^{6} \\ &=\left(\frac{7}{3}\right)^{-6} \end{aligned}$

(iii) $\begin{aligned}\left(\frac{5}{9}\right)^{-2} \times\left(\frac{5}{3}\right)^{2} \div\left(\frac{1}{5}\right)^{-2} &=\frac{\left(\frac{9}{5}\right)^{2} \times\left(\frac{5}{3}\right)^{2}}{\left(\frac{1}{5}\right)^{-2}} \\ &=\frac{9^{2}}{5^{2}} \times \frac{5^{2}}{3^{2}} \times \frac{1^{2}}{5^{2}} \\ &=3^{4-2} \times 5^{2-4} \\ &=3^{+2} \times 5^{-2}=5^{-2} \times \frac{1}{(3)^{-2}} \\ &=\left(\frac{5}{3}\right)^{-2} \end{aligned}$

(iv) $2^{-1}\left[\left(\frac{5}{3}\right)^{4}+\left(\frac{3}{5}\right)^{-2}\right] \div \frac{17}{9}$
=$\frac{1}{2}\left[\left(\frac{5}{3}\right)^{4}+\left(\frac{5}{3}\right)^{2}\right] \times \frac{9}{17}$
=$\frac{1}{2}\left[\left(\frac{5}{3}\right)^{2}\left[\left(\frac{5}{3}\right)^{2}+1\right]\right] \times \frac{9}{17}$

$=\frac{1}{2}\left[\frac{25}{9}\left(\frac{25}{9}+1\right)\right] \times \frac{9}{17}$

$=\frac{1}{2}\left[\frac{25}{9}\left(\frac{25+9}{9}\right)\right] \times \frac{9}{17}$

$=\frac{1}{2}\left[\frac{25}{9} \times \frac{34}{9}\right] \times \frac{9}{-17}$

$=\frac{1}{2}\left[\frac{25}{9} \times \frac{17 \times 2}{9} \times \frac{9}{17}\right]$

$=\frac{5^{2}}{3^{2}}=\left(\frac{5}{3}\right)^{2}$

$=\left(\frac{3}{5}\right)^{-2}$


(v) $(-7)^{3} \times\left(\frac{1}{-7}\right)^{-4} \div(-7)^{10}$
$=(-7)^{3} \times(-7)^{9} \div(-7)^{10}$

$=\frac{(-7)^{3} \times(-7)^{9}}{(-7)^{10}}$

$=(-7)^{3+9-10}$

$=(-7)^{2}=(7)^{2}$

$=\left(\frac{1}{(-7)}\right)^{-2}=\left(\frac{1}{7}\right)^{-2}$

$=\frac{1}{(+7)^{-2}}$

Question 9

Sol : 

(i) 
$\begin{aligned} \frac{49 \times Z^{-3}}{7^{-3} \times 10 \times Z^{-5}} &=\frac{7^{2} \times Z^{-3}}{7^{-3} \times 10 \times Z^{-5}} \\ &=\frac{7^{2+3} \times Z^{-3+5}}{10} \quad\left(\because \frac{a^{n}}{a^{n}}=a^{m-n}\right) \\ &=\frac{7^{5} \times Z^{2}}{10} . \end{aligned}$

(ii)
 $\begin{aligned} \frac{9^{3} \times 27 \times t^{4}}{(3)^{2} \times(3)^{4} \times t^{2}} &=\frac{\left(3^{2}\right)^{3} \times 3^{3} \times t^{4}}{(3)^{2} \times(3)^{4} \times t^{2}} \\ &=3^{6} \times 3^{3} \times t^{4} \times 3^{-2} \times 3^{-4} \times t^{-2} \\ &=3^{6+3-2-4} \times t^{4-2}\left(\because a^{m} \times a^{n}=a^{m+n}\right) \\ &=3^{3} \times t^{2} \\ &=27 \times t^{2} \end{aligned}$

(iii) $\frac{\left(3^{-2}\right)^{2} \times\left(5^{2}\right)^{-3} \times\left(t^{-3}\right)^{2}}{\left(3^{-2}\right)^{5} \times\left(5^{3}\right)^{-2} \times\left(t^{-4}\right)^{3}}=\frac{3^{-4} \times 5^{-6} \times t^{-6}}{3^{-10} \times 5^{-6} \times t^{-12}}\left(\therefore\left(a^{n}\right)^{n}=a^{m \times n}\right)$

$\begin{aligned} &=3^{-4+10} \times 5^{-6+6} \times t^{-6+12}\left(\because \frac{a^{m}}{a^{n}}=a^{m-n}\right) \\ &=3^{6} \times 5^{0} \times t^{6} \\ &=3^{6} \times t^{6} \quad\left(\therefore a^{0}=1\right) \end{aligned}$

(iv)
 $\begin{aligned} \frac{2^{-5} \times 15^{-5} \times 500}{5^{-6} \times 6^{-5}} &=\frac{2^{-5} \times(3 \times 5)^{-5} \times 5^{3} \times 2^{2}}{5^{-6} \times(2 \times 3)^{-5}} \\ &=\frac{2^{-5+2} \times 3^{-5} \times 5^{-5} \times 5^{3}}{5^{-6} \times 2^{-5} \times 3^{-5}} \\ &=2^{-5+2+5} \times 3^{-5+0+5} \times 5^{-5+6+3}=2^{2} \times 3 \times 5^{4} \\ &=2500 . \end{aligned}$

Question 10

Sol :
Let the number which should divide $\left(\frac{3}{-2}\right)^{-3}$ to got
$\left(\frac{2}{3}\right)^{2}$ be $x .$
$\Rightarrow \quad \frac{\left(\frac{3}{-2}\right)^{-3}}{x}=\left(\frac{2}{3}\right)^{2}$
$\Rightarrow \frac{3^{-3}}{(-2)^{-3}}=\frac{x \times 2^{2}}{3^{2}}$

Cross multiply
$\Rightarrow \quad 3^{-3} \times 3^{2}=x \times 2^{2} \times(-2)^{-3}$
$\Rightarrow \quad 3^{-3+2}=x \times(-2)^{2-3} \quad 3 \quad\left(\therefore(-a)^{2}=(a)^{2}\right)$
$\Rightarrow \quad 3^{-1}=x \times(-2)^{-1}$

$\begin{aligned} x &=\frac{3^{-1}}{(-2)^{-1}} \\ \therefore \quad x &=-2 / 3 \end{aligned}$

Question 11

Sol :
$\begin{aligned} & 9^{m} \div 3^{-2}=9^{4} . \\ \Rightarrow & \frac{9^{m}}{3^{-2}}=9^{4} \\ \Rightarrow & 9^{m} \times 9=9^{4} \quad\left(\because \frac{1}{a^{m}}=\right.\\ \Rightarrow & 9^{m+1}=9^{4} \\ & \text { If } \quad a^{m}=a^{n} \quad \text { then } \quad m=n \\ \Rightarrow & m+1=4 \end{aligned}$
     
$\therefore m=3$

Question 12

Sol :
$\left(\frac{-5}{7}\right)^{-4} \times\left(\frac{-5}{7}\right)^{12}=\left\{\left(\left(\frac{-5}{7}\right)^{3}\right)\right]^{x} \times\left(\frac{-5}{7}\right)^{-1}$
$\left(\frac{-5}{7}\right)^{-4+12}=\left(\frac{-5}{7}\right)^{3 x-1}$

$\Rightarrow \quad-4+12=3 x-1 \quad\left(\because a^{m}=a^{n} \Rightarrow m=n\right)$

$\Rightarrow \quad 3 x=9$

$\therefore x=3$

Question 13

Sol :
$\left(\frac{-2}{3}\right)^{-13} \times\left(\frac{3}{-2}\right)^{8}=\left(\frac{-2}{3}\right)^{-2 x+1}$
$\Rightarrow\left(\frac{-2}{3}\right)^{-13} \times\left(\frac{-2}{3}\right)^{-8}=\left(\frac{-2}{3}\right)^{-2 x+1} \quad\left(\because\left(\frac{p}{q}\right)^{m}=\left(\frac{q}{p}\right)^{-m}\right)$
$\Rightarrow\left(\frac{-2}{3}\right)^{-13-8}=\left(\frac{-2}{3}\right)^{-2 x+1}$

$\begin{aligned} \therefore-21=-2 x+1 &\left(\because a^{m}=a^{n} \text { then } m=n\right) \\ \Rightarrow &-2 x=-22 \\ \therefore x &=11  \end{aligned}$

Question 14

Sol :
(i) $5^{2 x-1}=\frac{1}{(125)^{x-3}} \Rightarrow 5^{2 x-1}=\frac{1}{\left(5^{3}\right)^{x-3}}$
$\Rightarrow \quad 5^{2 x-1} \times 5^{3(x-3)}=1$

$\Rightarrow \quad 5^{(2 x-1)+3(x-3)}=5 \quad\left(\because a^{0}=1\right)$

$\therefore \quad 2 x-1+3 x-9=0$

$\Rightarrow 5 x=10$

$\therefore x=2$


(ii) $\frac{9^{n} \times 3^{5} \times(27)^{3}}{3 \times(81)^{4}}=27$

$\Rightarrow \frac{\left(3^{2}\right)^{n} \times 3^{5} \times\left(3^{3}\right)^{3}}{3 \times\left(3^{4}\right)^{4}}=3^{3}$

$\Rightarrow 3^{(2 n+5+9)(-1-16)} \times 3^{3}$

$\Rightarrow 3^{2 n-3}=3^{3}$

$\therefore \quad 2 n-3=3$

$\Rightarrow n=3$

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

Image
 Exercise 15.1 Question 1 Using the given information, find the value of x in each of the following figures : Sol : (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° (Angles of a triangles) $\Rightarrow 42^{\circ}+x+50^{\circ}=180^{\circ}$ $\Rightarrow x+92^{\circ}=180^{\circ}$ $\Rightarrow x=180^{\circ}-92^{\circ}=88^{\circ}$ (ii) ∠ABD=∠ACD (Angles in the same segment) But ∠ACD=32° ∴∠ABD=32° Now in ΔABD ∠ABD+∠ADB+∠DAB=180° ⇒32°+45°+x=180° ⇒77°+x=180° ⇒x=180°-77°=103° (iii) ∠BAD=∠BCD (Angles in the same segement) But ∠BAD=20° $\therefore \angle B C D=20^{\circ}$ $\because \angle \mathrm{CEA}=90^{\circ}$ $\therefore \angle \mathrm{CED}=90^{\circ}$ Now in $\Delta$ CED, $\angle C E D+\angle B C D+\angle C D E=180^{\circ}$ $\Rightarrow 90^{\circ}+20^{\circ}+x=180^{\circ}$ $\Rightarrow 110^{\circ}+x=180^{\circ}$ $\Rightarrow x=180^{\circ}-110^{\circ}=70^{\circ}$ (iv) In ΔABC ∠ABC+∠ACB+∠BAC=180° (∵Sum of angles of a triangle) $\Rightarrow 69...
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.1

 Exercise 3.1 Question 1 (i) (2x+7y) 2 It is in the form of (a+b) 2 =a 2 +2ab+b 2 ∴a=2x, b=7y ∴(2x+7y) 2 =(2x) 2 +2.2x.7y+(7y) 2 =4x 2 +28xy+49y 2 (ii) $\left(\frac{1}{2}x+\frac{2}{3}y\right)^2$ Sol : $\left(\frac{1}{2}x \right)^2+2\times \frac{1}{2} \times x \times \frac{2}{3}\times y+\left(\frac{2}{3} y\right)^2$ $\frac{x^2}{4}+\frac{2xy}{3}+\frac{4}{9}y^2$ Question 2 (i) $\left(3x+\frac{1}{2x}\right)^2$ It is in the form of $(a+b)^2=a^2+2ab+b^2$ ∴$(3x)^2+2.3x.\frac{1}{2x}+\left(\frac{1}{2x}\right)^2$ $9x^3+3+\frac{1}{4x^2}$ (ii) $(3x^2y+5z)^2$ It is in the form of $(a+b)^2=a^2+2ab+b^2$ Here $a=3x^2y$ , b=5x $(3x^2y)^2+2.3x^2y.5z+(5z)^2$ $9x^4y^2+30x^2yz+25z^2$ Question 3 (i) $\left(3x-\frac{1}{2x}\right)^2$ It is in the form of $(a-b)^2=a^2-2ab+b^2$ Here, a=3x ; $b=\frac{1}{2x}$ $(3x)^2-2.3x.\frac{1}{2x}+\left(\frac{1}{2x}\right)^2$ $3^2-x^2-3+\frac{1}{2^2x^2}$ $9x^2-3+\frac{1}{4x^2}$ (ii) $\left(\frac{1}{2}x-\frac{3}{2}y\right)^2$ It is in the form of $(a-b)^2=a^2-2ab+b^2$ Here...
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S...
Read more