ML AGGARWAL CLASS 8 CHAPTER 10 Algebraic Expressions and Identities Exercise 10.5

  Exercise 10.5

Question 1

Sol :

(i) (3 x+5)(3 x+5)

$\begin{aligned}(3 x+5)^{2} &=(3 x)^{2}+2(3 x) .5+5^{2}\left(\therefore(a+b)^{2}=a^{2}+2 a b+b^{2}\right) \\ &=9 x^{2}+30 x+25 \end{aligned}$

(ii) $\begin{aligned}(9 y-5) &(9 y-5) \\(9 y-5)^{2} &=(9 y)^{2}-2(9 y) 5+(+5)^{2}\left(\because(a-b)^{2}=a^{2}-2 a b+b^{2}\right) \\ &=81 y^{2}-90 y+25 \end{aligned}$

(iii) $(4 x+11 y)(4 x-11 y)$

=$(4 x)^{2}-(11 y)^{2}\left(\because(a+b)(a-b)=a^{2}-b^{2}\right)$

=$16 x^{2}-121 y^{2}$

(v) $\left(\frac{2}{a}+\frac{5}{b}\right)\left(\frac{2}{a}+\frac{5}{b}\right)$

$\begin{aligned}\left(\frac{2}{a}+\frac{5}{b}\right)^{2} &=\left(\frac{2}{a}\right)^{2}+2 \cdot \frac{2}{a} \cdot \frac{5}{b}+\left(\frac{5}{b}\right)^{2}\left(\because \cdot(a+b)^{2}=a^{2}+2 a b+b^{2}\right) \\ &=\frac{4}{a^{2}}+\frac{20}{a b}+\frac{25}{b^{2}} \end{aligned}$

(vi) $\left(\frac{p^{2}}{2}+\frac{2}{q^{2}}\right)\left(\frac{p^{2}}{2}-\frac{2}{q^{2}}\right)$

$\left(\frac{p^{2}}{2}\right)^{2}-\left(\frac{2}{q^{2}}\right)^{2} \quad\left(\because(a+b)(a-b)=a^{2}-b^{2}\right.$

$\frac{p^{4}}{4}-\frac{4}{q^{4}}$

Question 2

Sol :

(i) $\begin{aligned} 81^{2} &=(80+1)^{2} \\ &=80^{2}+2.80 .1+1^{2}\left(\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right) \\ &=6400+160+1 \\ 81^{2} &=6561 \end{aligned}$

(ii) 

$\begin{aligned} 97^{2} &=(100-3)^{2} \\ &=(100)^{2}-2 \cdot 100 \cdot 3+3^{2}\left(\because(a-b)^{2}=a^{2}-2 a b+b^{2}\right] \\ &=10000-600+9 \\ &=9409^{} \end{aligned}$

(iii)

 $\begin{aligned} 105^{2} &=(100+5)^{2} \\ &=100^{2}+2.100 \cdot 5+5^{2}\left(\cdots(a+b)^{2}=a^{2}+2 a b+b^{2}\right) \\ &=10000+1000+25 \\ &=11025 \end{aligned}$

(iv) 

$\begin{aligned} 997^{2} &=(1000-3)^{2} \\ &=(1000)^{2}-2 \cdot 1000 \cdot 3+3^{2}\left(\because(a-b)^{2}=a^{2}-2 a b+b^{2}\right) \\ &=1000000-6000+9 \\ 997^{2} &=994009 \end{aligned}$

(v) 

$\begin{aligned} 6.1^{2} &=(6+0.1)^{2} \\ &=6^{2}+2 \cdot 6 \cdot(0.1)+0.1^{2}\left(\because(a+b)^{\gamma}=a^{2}+2 a b+b^{2}\right) \\ &=36+1.2+0.01 \\ 6.1^{2} &=31.21 \end{aligned}$

(vi) 

$\begin{aligned} 496 \times 504 &=(500-4)(500+4) \\ &=(500)^{2}-4^{2}\left(\because \cdot(a+b)(a-b)=a^{2}-b^{2}\right) \\ &=250000-16 \\ 496 \times 504 &=249984 \end{aligned}$

(vii)

 $\begin{aligned} 20.5 \times 19.5 &=(20+0.5)(20-0.5) \\ & \left.=20^{2}-0.5^{2}(19+b)(a-b)=a^{2}-b^{2}\right) \\ &=400-0.25 \\ 20.5 \times 19.5 &=399.75 \end{aligned}$

(viii) 

$\begin{aligned} 9.6^{2} &=(10-0.4)^{2} \\ &=10^{2}-2.10 \cdot(0.4)+(0.4)^{2}\left(\because \cdot(a-b)^{2}=a^{2}-2 a b+b^{2}\right) \\ &=100-8+0.16 \\ 9.6^{2} &=92.16 \end{aligned}$

Question 3

Sol :

(i) 

$\begin{aligned}(p q+5 r)^{2} &=(P q)^{2}+2 \cdot p q \cdot 5 r+(5 r)^{2}\left(-(a+b)^{2}=a^{2}+b^{2}+2 a b\right) \\ &=p^{2} q^{r}+10 p q r+25 2^{2} \end{aligned}$

(ii) 

$\begin{aligned}\left(\frac{5}{2} a-\frac{3}{5} \cdot b\right)^{2} &=\left(\frac{5}{2} \cdot a\right)^{2}-2 \cdot \frac{5}{2} \cdot a \cdot \frac{3}{5} \cdot b+\left(\frac{3}{5} b\right)^{2} \\ &\left(\because(a-b)^{2}=a^{2}-2 a b+b^{2}\right) \\ &=\frac{25}{4} a^{2}-3 a b+\frac{9}{25} b^{2} \end{aligned}$

(iii) 

$\begin{aligned}(\sqrt{2} \cdot a+\sqrt{3} \cdot b)^{2} &=(\sqrt{2} \cdot a)^{2}+2 \sqrt{2} \cdot a \cdot \sqrt{3} \cdot b+(\sqrt{3} \cdot b)^{2} \\(\therefore& \left.(a+b)^{2}=a^{2}+2 a b+b^{2}\right) \\ &=2 a^{2}+2 \sqrt{6} a b+3 b^{2} \end{aligned}$

(iv) 

$\left(\frac{2 x}{3 y}-\frac{3 y}{2 x}\right)^{2}=\left(\frac{2 x}{3 y}\right)^{2}-2 \cdot \frac{2 x}{3 y} \cdot \frac{3 y}{2 x}+\left(\frac{3 y}{2 x}\right)^{2}$ $\left(∴(a+b)^{2}=a^{2}+b^{2}+2 a b\right)$

$=\frac{4 x^{2}}{9 y^{2}}-2+\frac{9 y^{2}}{4 x^{2}}$

Question 4

Sol :

(i) 

$\begin{aligned}(x+7)(x+3) &=x^{2}+(7+3) x+7 \times 3\left(:(x+a)(x+b)=x^{2}+(a+b) x+a b\right) \\ &=x^{2}+10 x+21 \end{aligned}$

(ii) $(3 x+4)(3 x-5)=(3 x)^{v}+(4+(-5))(3 x)+4 x-5$

$\left(\because(x+a)(x+b)=x^{2}+(a+b) x+a b\right)$

$=9 x^{2}-3 x-20$

(iii) $\left(p^{2}+2 q\right)\left(p^{2}-3 q\right)=\left(p^{2}\right)^{2}+(2 q+(-3 q)) p^{2}+2 q x-3 q$

               

$\left(\because(x+a)(x+b)=x^{2}+(a+b) x+a b\right)$

$=p^{4}-^{2} q^{2}-6 q^{2}$

$=p^{4}-p_{q}^{2}-6 q^{2}$

(iv) $(a b c+3)(a b c-5)=(a b c)^{2}+(3+(-5)) \cdot a b c+3 x-5$

$\left(\because(x+a)(x+b)=x^{2}+(a+b) \cdot x+a b\right)$

$=(a b c)^{2}-2 a b c-15$

 

Question 5

Sol :

(i) $203 \times 204=(200+3)(200+4)$

$=(200)^{2}+(3+4) 200+3 \times 4\left(\because(x+a)(x+b)=x^{2}+(a+b)+x+ab\right.$

=40000+1400+12

=41412

(ii) $8.2 \times 8.7=(8+0.2)(8+0.7)$

$=8^{2}+(0.2+0.7) 8+0.2 \times 0.7$

=64+7.2+0.14

=71.34

(iii) $107 \times 93=(100+7)(100-7)$

$=(100)^{2}+(7+(-7)) \cdot 100+7 x-7$

$\left(\because(x+a)(x+b)=x^{2}+(a+b) \cdot x+a b\right)$

=10000+0.100=49

=9951

Question 6 

Sol :

(i) $53^{2}-47^{2}=(53+47)(53-47)$ $\left(∴ a^{2}-b^{2}=(a+b)(a-b)\right)$

=(100)(6)

=600

(ii) $(2.05)^{2}-(0.95)^{2}=(2.05+0.95)(2.05-0.95)$

$=3 \times 0.1$

=0.3

Question 7

Sol :

i. $(2 x+5 y)^{2}+(2 x-5 y)^{2}$

$(2 x)^{2}+(5 y)^{2}+2 \cdot 2 x \cdot 5 y+(2 x)^{2}+(5 y)^{2}-2 \cdot 2 x \cdot 5 y$

$\because(a+b)^{2}=a^{2}+b^{2}+2 a b$

$2(2 x)^{2}+2(5 y)^{2}$

$2 \cdot\left[4 x^{-1}+25 y^{2}\right]$

$8 x^{2}+50 y^{2}$

                                

(ii) $\left(\frac{7}{2} a-\frac{5}{2} b\right)^{2}-\left(\frac{5}{2} a-\frac{z}{2} b\right)^{2}$

=$\left(\frac{7}{2} a\right)^{2}+\left(\frac{5}{2} b\right)^{2}-2 \cdot \frac{7}{2} a \cdot \frac{5}{2} \cdot b-\left[\left(\frac{5}{2} a\right)^{n}+\left(\frac{7}{2} b\right)^{2}-2 \cdot \frac{5}{2} \cdot a \cdot \frac{7}{2} b\right]$

=$\frac{49}{4} a^{2}+\frac{25}{4} b^{2}-2 \cdot \frac{7}{2} \cdot \frac{5}{2} \cdot b-\frac{25}{4} a^{2}-\frac{49}{4} b^{2}+2 \cdot \frac{5}{2} a\cdot \frac{7}{2} \cdot b$

=$\left(\frac{49}{4}-\frac{25}{4}\right) a^{2}+\left(\frac{25}{4}-\frac{49}{4}\right) b^{2}$

=$\frac{24}{4} \cdot a^{2}-\frac{24}{4} \cdot b^{2}$

=$6\left(a^{2}-b^{2}\right)$

(iii) $\left(p^{2}-qr^{2} \cdot \right)^{2}+2 p^{2} q^{2} 2$

$\left(p^{2}\right)^{2}-2 \cdot p^{2} \cdot q^{2} \cdot 2+\left(q^{2} \cdot 2\right)^{2}+2 p^{2} q^{2} \cdot r$   $\left(\therefore(a+b)^{2}=a^{2}+2 a b+b^{2}\right.$

$p^{4}-2 p^{2} q^{2} \cdot r+q^{4} \cdot r^{2}+2 p^{2} q^{2} \cdot r$

$p^{4}+q^{4} \cdot r^{2}$

Question 8

Sol :

L H S

(i) $(4 x+7 y)^{2}-(4 x-7 y)^{n}$

$(4 x)^{2}+(7 y)^{2}+2 \cdot 4 x \cdot 7 y-\left[(4 x)^{2}+(7 y)^{2}-2.4 x-7 y\right]$

$\left(\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right)$

$(4 x)^{2}+(7 y)^{2}+2.4 x \cdot 7 y-(4 x)^{2}-(7 y)^{2}+2 \cdot 4 x \cdot 7 y$

$4.4 x \cdot 7 y$

112 x y= R.H.S

(ii) $\left(\frac{3}{7} p-\frac{7}{6} q\right)^{2}+p q$

$\left(\frac{3}{7} p\right)^{2}+\left(\frac{7}{6} q\right)^{2}-2 \cdot \frac{3}{7} p \cdot \frac{1}{6} \cdot q+p q$

$\left(\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right)$

$\frac{9}{49} p^{2}+\frac{49}{36} q^{2}-p q+p q$

$\begin{aligned} \frac{9}{49} \cdot p^{2}+\frac{49}{36} q^{2} &=R \cdot H \cdot S \\ \therefore \quad L . H \cdot S &=R \cdot H \cdot S \end{aligned}$

(iii) $L \cdot H \cdot s=(p-q)(p+q)+(p-r)(q+r)+(r-p)(r+p)$

$=p^{2}-q^{2}+q^{2}-r^{2}+r^{2}-p^{2}$

$\left(\because+(a+b)(a-b)=a^{2}-b^{2}\right]$

=0=RHS

$\therefore \quad L H S=R H S$

Question 9

Sol :

Given : $\left(x+\frac{1}{x}\right)=2$

Squaring on both sides

(i) $\left(x+\frac{1}{x}\right)^{2}=2^{2}$

$x^{2}+2 \cdot x \cdot \frac{1}{x}+\left(\frac{1}{x}\right)^{2}=$$4\left(\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right)$

$x^{2}+2+\frac{1}{x^{2}}=4$

$x^{2}+\frac{1}{x^{2}}=4-2$

$x^{2}+\frac{1}{x^{2}}=2$

(ii) Again squaring on both sides

$\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=2^{2}$

$\left(x^{2}\right)^{n}+2 \cdot x^{2}-\frac{1}{x^{2}}+\left(\frac{1}{x^{2}}\right)^{2}=4$

$x^{4}+2+\frac{1}{x^{4}}=4$

$x^{4}+\frac{1}{x^{4}}=4-2$

$x^{4}+\frac{1}{x^{4}}=2$n

Question 10

Sol :

(i) $x-\frac{1}{x}=7$

Squaring on both sides

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

$x^{2}-2 \cdot x \cdot \frac{1}{x}+\left(\frac{1}{x}\right)^{2}=49$

$\left(\because(a-b)^{2}=a^{2}-2 a b+b^{2}\right)$

$x^{2}-2+\frac{1}{x^{2}}=49$

$x^{2}+\frac{1}{x^{2}}=49+2$

$x^{2}+\frac{1}{x^{2}}=51$

(ii) $x^{2}+\frac{1}{x^{2}}=51$

squaring on both sides

$\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=51^{2}$

$\left(x^{2}\right)^{2}+2 \cdot x^{2} \cdot \frac{1}{x^{2}}+\left(\frac{1}{x^{2}}\right)^{2}=2601$

$x^{4}+2+\frac{1}{x^{4}}=2601$

$x^{4}+\frac{1}{x^{4}}=2601-2$

$x^{4}+\frac{1}{x^{4}}=2599$

Question 11

Sol :

$x^{2}+\frac{1}{x^{2}}=23$

(i) $x^{2}+\frac{1}{x^{2}}=23$

Adding 2 on both sides

$x^{2}+\frac{1}{x^{2}}+2=23+2$

$(x)^{2}+\left(\frac{1}{x}\right)^{2}+2 \cdot x \cdot \frac{1}{x}=25\left(\because a^{2}+b^{2}+2 a b=(a+b)^{2}\right)$

$\left(x+\frac{1}{x}\right)^{2}=25$

$x+\frac{1}{x}=5$

(ii)  $x+\frac{1}{x^{2}}=25^{2}$

Subtract '2' on both sides

$x^{2}+\frac{1}{x^{2}}-2=23-2$

$(x)^{2}+\left(\frac{1}{x}\right)^{n}-2 \cdot x \cdot \frac{1}{x}=21$

$\left(x-\frac{1}{x}\right)^{2}=21\left(\because \quad a^{2}+b^{2}-2 a b=(a-b)^{2}\right)$

$x-\frac{1}{x}=\sqrt{21}$

$x-\frac{1}{x}=3 \sqrt{3}$

Question 12

Sol :

given a+b=9,  a b=10

Squaring on both sides

$(a+b)^{2}=9^{2}$

$a^{2}+b^{2}+2 a b=81$

$a^{2}+b^{2}+2 \times 10=81 \quad(\because$ given $a b=10)$

$a^{2}+b^{2}+20=81$

$a^{2}+b^{2}=61$

Question 13

Sol :

given $a-b=6, a^{2}+b^{2}=42$

a-b=6

Squaring on both Sides

$(a-b)^{2}=6^{2}$

$a^{2}+b^{2}-2 a b=36$

$42-2 a b=36\left(\because a^{2}+b^{2}=42\right)$

42-36=2 a b

2 a b=6

a b=3

Question 14

Sol :

given $a^{2}+b^{2}=41, a b=4$

(i) Consider

$\begin{aligned}(a+b)^{2} &=a^{2}+b^{2}+2 a b \\(a+b)^{2} &=41+2 \times u=41+8 \\(a+b)^{2} &=49 \\ a+b &=7 \end{aligned}$

(ii) Consider 

$\begin{aligned}(a-b)^{2} &=a^{2}+b^{2}-2 a b \\ &=41-2 \times 4=41-8=33 . \\(a-b)^{2} &=33 \\ a-b &=\sqrt{33} \end{aligned}$