SChand CLASS 9 Chapter 3 Expansions Exercise 3(B)

 Exercise 3(B)

Question 1

Find the value of:

(i) a² + b² when a + b = 9, ab = 20

(ii) p² + q² if p – q = 6 and p + q = 14

(iii) mn if m + n = 8, m – n = 2

(iv) $x^2 + \frac{1}{x^2}$ and $x^4 + \frac{1}{x^4}$ if $x + \frac{1}{x}= 3$

(v) $x – \frac{1}{x}$ and $x^2 – \frac{1}{x^2}$ if $x + \frac{1}{x} = \sqrt{5}$

Sol :

(i) a+b=9  , ab=20

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

$\Rightarrow \quad(9)^{2}-2 \times 20$

⇒81-40

⇒41

(ii) p-q=6 , p+q=14

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

⇒$(14)^{2}+(6)^{2}$

⇒196+36

⇒232

∴$p^{2}+q^{2}= \frac{232}{2}$

=116

(iii) m+n=8 , m-n=2

$\begin{aligned} 4 m n & \Rightarrow(m+n)^{2}-(m-n)^{2} \\ &=\left(81^{2}-(2)^{2}\right.\end{aligned}$

=64-4=60

∴mn$=\frac{60}{4}=15$

(iv) $x+\frac{1}{x}=3$

Squaring both sides

$\left(x-\frac{1}{x}\right)^{2}=(3)^{2}$

$x^{2}+\left(\frac{1}{x}\right)^{2}+2 \times x \times \frac{1}{y} \Rightarrow 9$

$x^{2}+\frac{1}{x^{2}}+2 \Rightarrow 9$

$x^{2}+\frac{1}{x^{2}} \Rightarrow 9-2$

$x^{2}+\frac{1}{x^{2}} \Rightarrow 7$

Again , Squaring both sides

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

$x^{4}+\frac{1}{x^{4}}+2 \times x^{2} \times \frac{1}{x^{2}} \Rightarrow 49$

$x^{4}+\frac{1}{x^{4}}+2 \Rightarrow 49$

$x^{4}+\frac{1}{x^{4}}=47$

(v) $x+\frac{1}{x} \Rightarrow \sqrt{5}$

Squaring both sides

$\left(x+\frac{1}{x}\right)^{2}=(\sqrt{5})^{2}$

$x^{2}+\left(\frac{1}{x}\right)^{2}+2 \times x \times \frac{1}{x} \Rightarrow 5$

$x^{2}+\frac{1}{x^{2}}+2 \Rightarrow 5$

$x^{2}+\frac{1}{x^{2}} \Rightarrow 3$

Now , $\left(x-\frac{1}{x}\right)^{2} \Rightarrow x^{2}+\frac{1}{x^{2}}-2$

=3-2=1

$\left(x-\frac{1}{x}\right)^{2} \Rightarrow(\pm 1)^{2}$

$x-\frac{1}{x}=\pm 1$

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

$x^{2}-\frac{1}{x^{2}}=15 \times(\pm 1)$

$x^{2}-\frac{1}{x^{2}} \Rightarrow \pm \sqrt{5}$

Question 2

(i) a² + b² + c² if a + b + c = 17 and ab + bc + ca = 30.

(ii) ab + be + ca, a + b + c = 15 and a² + b² + c² = 77

(iii) a + b + c if a² + b² + c² = 50 and ab + bc + ca =47.

Sol :

(i) a+b+c=17

Squaring both sides

$(a+b+c)^{2} \Rightarrow(17)^{2}$

$a^{2}+b^{2}+c^{2}+2(a b+b c+c a) \Rightarrow 289$

$a^{2}+b^{2}+c^{2}+2 \times 30 \Rightarrow 289$  [∵ab+bc+ca=30]

$a^{2}+b^{2}+c^{2} \Rightarrow 289-60$

$a^{2}+b^{2}+c^{2}=229$

(ii) a+b+c=15

$a^{2}+b^{2}+c^{2}=77$

a+b+c=15

Squaring both sides

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

$a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=525$

77+2(ab+bc+ca)=925

2(ab+bc+ca)=25-77

2(ab+bc+ca)=148

ab+bc+ca$=\frac{148}{2}$

ab+bc+ca=74

(iii) $a^{2}+b^{2}+c^{2}=50$ and ab+bc+ca=47

$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$

=50+2×47

=50+94=144

$(a+b+c)^2=(\pm 12)^{2}$

a+b+c=土12

Question 3

(i) 8x³ + 84x²y + 294xy² + 343y³ if x = 1, y = 2

(ii) 27x³ – 27x²y + 9xy2 – y³ if x = 2, y = 1

Sol :

(i) $8 x^{3}+84 x^{2} y+294 x y^{2}+343 y^{3}$ if x=1, y=2

$8 x^{3}+84 x^{2} y+294 x y^{2}+343 y^{3}$

$(2 x)^{3}+3(2 x)^{2} \times 7 y+3 \times 2 x \times(7 y)^{2}+(7 y)^{3}$

$(2 x)^{3}+3(2 x)^{2} \times 7 y+3 \times 2 x \times(7 y)^{2}+(7 y)^{3}$

$(2 x+7 y)^{3}=(2 \times 1+7 \times 2)^{3}$

$(2+14)^{3}$

$(16)^{3}$

=4096

(ii) x=2 , y=1

$27 x^{3}-27 x^{2} y+9 x y^{2}-y^{3}$

$(3 x)^{3}-3(3 x)^{2} y+3(3 x) y^{2}-(y)^{3}$

$(3 x-y)^{3}=\left(3 \times 2-1\right)^{3}$

$(6-1)^{3}$

$(5)^{3}=125$

Question 4

(i) a³ + b³ if a + b = 3 and ab = 2

(ii) $x^3+\frac{1}{x^3}$ if $\left(x+\frac{1}{x}\right)=3$

(iii) $x^3-\frac{1}{x^3}$ if $\left(x^2+\frac{1}{x^2}\right)=18$

(iv) $x^3+\frac{1}{125x^3}$ if $x^2+\frac{1}{25x^2}=8\frac{3}{5}$

Sol :

(i) a+b=3  ab=2

$a^{3}+b^{3}=(a+b)^{3}-3 a b(a-b)$

$=(3)^{3}-3 \times 2 \times 3$

=27-18=9

(ii) $x+\frac{1}{x}=3$

$x^{3}+\frac{1}{x^{3}} \Rightarrow\left(x+\frac{1}{x}\right)^{3}-3\left(x+\frac{1}{x}\right)$

$(3)^{3}-3(3)$

=27-9=18

(iii) $x^{2}+\frac{1}{x^{2}} \Rightarrow 18$

(-2) From both sides

$x^{2}+\frac{1}{x^{2}}-2 \Rightarrow 18-2$

$\Rightarrow\left(x-\frac{1}{x}\right)^{2}=16$

$\Rightarrow\left(x-\frac{1}{x}\right)^{2}=(\pm 4)^{2}$

$x-\frac{1}{x}=\pm 4$

Now $x^{3}-\frac{1}{x^{3}}=\left(x-\frac{1}{x}\right)^{3}+3\left(x-\frac{1}{x}\right)$

when $x-\frac{1}{x} \Rightarrow 4$

⇒$(4)^{3}+3 \times 4$

⇒64+12=76

and $x^{3}-\frac{1}{x^{3}} \Rightarrow\left(x+\frac{1}{x}\right)^{3}+3\left(x-\frac{1}{x}\right)$

⇒$(-4)^{3}+3 \times (-4)$

-64-12=-76

∴$x^{3}-\frac{1}{x^{3}} \Rightarrow \pm 76$

and 

$x^{3}-\frac{1}{x^{3}} \Rightarrow\left(x+\frac{1}{x}\right)^{3}+3\left(x-\frac{1}{x}\right)$

$(-4)^{3}+3 \times(-4)$

=-64-12=-76

∴$x^{3}-\frac{1}{x^{3}} \Rightarrow \pm 76$

(iv) $x^{2}+\frac{1}{25 x^{2}}=8 \frac{3}{5} \Rightarrow \frac{43}{5}$

$x^{2}+\left(\frac{1}{5 x}\right)^{2}=\frac{43}{5}$

Adding both sides $\frac{2}{5}$ $\left[2 \times x+\frac{1}{5 x} \rightarrow \frac{2}{5}\right)$

$x^{2}+\frac{1}{(5 x)^{2}}+\frac{2}{5}=\frac{43}{5}+\frac{2}{5}$

$\Rightarrow\left(x+\frac{1}{57}\right)^{2}=\frac{43+2}{5}$

$\Rightarrow\left(x+\frac{1}{5 x}\right)^{2} \Rightarrow \frac{45}{5}$

$\Rightarrow\left(x+\frac{1}{5 x}\right)^{2} \Rightarrow 9$

$7\left(x+\frac{1}{5 x}\right)^{x}=(\pm 3)^{2}$

$x+\frac{1}{5 x}=\pm 3$

When $x+\frac{1}{5 x} \Rightarrow 3$

then $x^{3}+\frac{1}{125 x^{3}}=\left(x+\frac{1}{5 x}\right)^{3}-3 \times x+\frac{1}{5 x}\left(x+\frac{1}{5 x}\right)$

⇒$(3)^{3}-\frac{3}{5} \times 3$

⇒$27-\frac{9}{5}$

⇒$\frac{135-9}{5}$

⇒$\frac{126}{5} \Rightarrow 25 \frac{1}{5}$

When  $x+\frac{1}{5 x}\Rightarrow -3$

then 

$x^{3}+\frac{1}{125 x^{3}}=\left(x+\frac{1}{5 x}\right)^{3}-3 x \times \frac{1}{5 x}\left(x+\frac{1}{5x} \right)$

$=(-3)^{3}-\frac{3}{5} \times(-3)$

$=-27+\frac{9}{5}$

$=\frac{-135+9}{5}$

$=\frac{-126}{5}$

$=-25 \frac{1}{5}$

Hence , $x^{3}+\frac{1}{25 x^{3}}=\pm 25 \frac{1}{5}$

Question 5

Evaluate:

(i) 102 × 98

(ii) 1003² – 997²

(iii) (10)³ – (5)³ – (5)³

Sol :

(i) 10² × 98

= (100 + 2) (100-2) {∵ {a + b)(a – b) = a² – b²)}

= (100)² – (2)²

= 10000 – 4

= 9996

(ii) (1003)² – (997)²

= (1003 + 997) (1003 – 997) {∵ a² – b² = (a + b) (a – b)}

= 2000 × 6 = 12000

(iii) (10)³ – (5)³ – (5)³ = (10)³ + (- 5)³ + (- 5)³

Let 10 = a, – 5 = b, – 5 = c and a + b + c – 10 – 5 – 5 = 0

∵ a + b + c = 0. then a³ + b³ + c³ = 3abc

⇒ (10)³ – (5)³ – (5)³ = 3 × 10 × (- 5) (- 5) = 750