Question

if x² +(1/x)²= 7,
then find the value of x² - (1/x)²


source : advisor textbook ( class 8 )


__help this kid plz¡

Answer 1

Answer :

x² - 1/x² = 3√5

$\rule{150}2$

Given:-

• $\sf{x^2\:+\: \bigg(\dfrac{1}{x} \bigg)^{2} \:=\:7}$

Find:-

$\sf{x^2\:-\: \bigg(\dfrac{1}{x} \bigg)^{2}}$

Solution:-

$\implies\:\sf{x^2\:+\:\dfrac{1}{x^2}\:=\:7}$

$\implies\:\sf{x^2\:+\:\dfrac{1}{x^2}\:=\:9\:-\:2}$

$\implies\:\sf{x^2\:+\:\dfrac{1}{x^2}\:+\:2\:=\:9}$

(a + b)² = a² + b² + 2ab

$\implies\:\sf{\bigg(x\:+\:\dfrac{1}{x} \bigg)^2\:=\:9}$

$\implies\:\sf{x\:+\:\dfrac{1}{x}\:=\:\sqrt{9}}$

$\implies\:\sf{x\:+\:\dfrac{1}{x}\:=\:3}$ ....(1)

We have to find $\sf{x^2\:-\:\dfrac{1}{x^2}}$

$\implies\:\sf{x^2\:+\:\dfrac{1}{x^2}\:-\:\dfrac{2x}{x}=\:7\:-\:2}$

$\implies\:\sf{\bigg(x\:-\:\dfrac{1}{x}\bigg)^2\:=\:5}$

$\implies\:\sf{x\:-\:\dfrac{1}{x}\:=\:\sqrt{5}}$ ....(2)

(a + b) (a - b) = a² - b²

$\implies\:\sf{\bigg(x\:+\:\dfrac{1}{x}\bigg)\bigg(x\:-\:\dfrac{1}{x}\bigg)\:=\:3\sqrt{5}}$

$\implies\:\sf{x^2\:-\:\dfrac{1}{x^2}\:=\:3\sqrt{5}}$

Answer 2

Solution :-

Given :-

x² + ( 1/x )² = 7

To solve this question we can think of the identity a² - b² = (a + b)(a - b)

Here according to the question a = x, b = 1/x

So, find the (a + b) i.e x + ( 1/x ) and (a - b) i.e x - ( 1/x )

Finding the value of (x + 1/x)

$\tt {x}^{2} + \bigg( \dfrac{1}{x} \bigg)^{2} = 7$

Adding 2(x)(1/x) on both sides

$\tt \implies {x}^{2} + \bigg( \dfrac{1}{x} \bigg)^{2} + 2(x) \bigg( \dfrac{1}{x} \bigg) = 7 + 2(x) \bigg( \dfrac{1}{x} \bigg)$

$\tt \implies \bigg(x + \dfrac{1}{x} \bigg)^{2} =9$

[ Because, a² + b² + 2ab = (a + b)² ]

$\tt \implies x + \dfrac{1}{x} = \sqrt{9}$

$\tt \implies x + \dfrac{1}{x} = 3$

Finding the value of (a - b) i.e x - ( 1/x )

$\tt {x}^{2} + \bigg( \dfrac{1}{x} \bigg)^{2} = 7$

Subtracting 2(x)(1/x) on both sides

$\tt \implies {x}^{2} + \bigg( \dfrac{1}{x} \bigg)^{2} - 2(x) \bigg( \dfrac{1}{x} \bigg) = 7 - 2(x) \bigg( \dfrac{1}{x} \bigg)$

$\tt \implies {x}^{2} + \bigg( \dfrac{1}{x} \bigg)^{2} - 2(x) \bigg( \dfrac{1}{x} \bigg) = 7 - 2$

$\tt \implies \bigg( x - \dfrac{1}{x} \bigg)^{2} = 5$

[ Because, a² + b² - 2ab = (a - b)² ]

$\tt \implies x - \dfrac{1}{x} = \sqrt{5}$

Now, finding x² - ( 1/x )²

We know that a² - b² = (a + b)(a - b)

Now substituting the values

$\tt \implies {x}^{2} - \bigg( \dfrac{1}{x} \bigg)^{2} = 3 \sqrt{5}$

Hence, the value of x² - ( 1/x )² is 3√5.