Question

if x -1/x = 5 , find x^2 + 1/x^2​

Answer 1

Step-by-step explanation:

Given that:

$\tt{x - \frac{1}{x} = 5}$

To find:

$\tt{The \: value \: of \: {x}^{2} + \frac{1}{ {x}^{2} } }$

Solution:

We have,

$\tt{x - \frac{1}{x} = 5}$

Squaring on both sides, we get

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

Using algebraic Identity

$\tt{(x - y {)}^{2} } = (x - y)(x - y) = {x}^{2} - 2xy + {y}^{2} ,we \: get$

________________________

Now,

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

$⟹ \tt{ {x}^{2} - 2x \times \frac{1}{x} + \bigg( \frac{1}{x} \bigg)^{2} = 25}$

$⟹ \tt{{x}^{2} - 2 \cancel{x }\times \frac{1}{ \cancel{x}} + \bigg( \frac{1}{x} \bigg)^{2} = 25 }$

$⟹ \tt{ {x}^{2} - 2 + \frac{1 }{ {x}^{2} } } = 25$

$⟹ \tt{ { x}^{2} + \frac{1}{ {x}^{2} } = 25 + 2}$

$⟹ \tt{{x}^{2} + \frac{1}{ {x}^{2} } = 27}$

Answer:-

$\tt \red{Hence,the \: value \: of \: \: {x}^{2} + \frac{1}{ {x}^{2} } = 27}$

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