Question

Solve the Problem:-
$If \: x - \frac{1}{x} \: = 3, \: find \: the \: value \: of \: {x}^{2} + \frac{1}{ {x}^{2} } \: and \: {x}^{4} + \frac{1}{ {x}^{4} }$
Chapter- Algebraic Expressions.

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Answer 1

Answer:

Your answer attached in the photo

Answer 2

${\huge{\underline{\underline{\rm{Question}}}}}$

★ If $\rm x-\dfrac{1}{x}=3$ find the value of $\rm x^2 + \dfrac{1}{x^2}$ and $\rm x^4 + \dfrac{1}{x^4}$.

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★ Given :

• $\rm \: x - \dfrac{1}{x} = 3.....(i)$

★ To Find :

1. $\rm \: {x}^{2} + \dfrac{1}{ {x}^{2} }$
2. $\rm \: {x}^{4} + \frac{1}{ {x}^{4} }$

★ Solution :

# 1st one ...

Squaring both sides of eq(i)

$\mapsto \rm \: {(x - \dfrac{1}{x}) }^{2} = {3}^{2}$

$\mapsto \rm \: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 \times \cancel x \times \dfrac{1}{ \cancel x}$

$\mapsto \rm \: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 = 9$

$\mapsto \rm \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 9 + 2$

$\mapsto \boxed{ \rm \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 11}....(ii)$

# 2nd one ..

Squaring both sides of eq(ii)

$\mapsto \rm \: {( {x}^{2} + \dfrac{1}{ {x}^{2} } ) }^{2} = {11}^{2}$

$\mapsto \rm \: {x}^{4} + \dfrac{1}{ {x}^{4} } + 2 \times \cancel{{x}^{2} } \times \dfrac{1}{ \cancel{{x}^{2} }} = 121$

$\mapsto \rm \: {x}^{4} + \dfrac{1}{ {x}^{4} } + 2 = 121$

$\mapsto \rm \: {x}^{4} + \dfrac{1}{ {x}^{4} } = 121 - 2$

$\mapsto \boxed{\rm \: {x}^{4} + \dfrac{1}{ {x}^{4} } = 119}$

We have found the answers of both questions ..

★ Formula Used :

• $(a+b)^2 = a^2 + b^2 + 2ab$
• $(a-b)^2 = a^2 + b^2 - 2ab$