Question

if x - 1/x =5 find the value of x square+1/x square and xsquare+1/x to the power four​

Answer 1

Solution:-

Given:-

$\rm \: x - \frac{1}{x} = 5$

To find value of

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

Now take

$\rm \: x - \frac{1}{x} = 5$

squaring on both side

$\rm \:( x - \frac{1}{x} ) {}^{2} =( 5) {}^{2}$

Use this identity

$\rm \: (a - b) {}^{2} = {a}^{2} + {b}^{2} - 2ab$

We get

$\rm {x}^{2} + \frac{1}{ {x}^{2} } - 2 \times x \times \frac{1}{x} = 25$

$\rm {x}^{2} + \frac{1}{ {x}^{2} } - 2 \times \not x \times \frac{1}{ \not x} = 25$

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

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

$\boxed{ \rm \: \rm {x}^{2} + \frac{1}{ {x}^{2} } = 27}$

Now take

${ \rm \: \rm {x}^{2} + \frac{1}{ {x}^{2} } = 27}$

Squaring on both side

$\rm \: ( {x}^{2} + \frac{1}{ {x}^{2} } ) {}^{2} = (27) {}^{2}$

we get ,

$\rm \: ( {x}^{2} ) {}^{2} + \frac{1}{( {x}^{2} ) {}^{2} } + 2 \times x {}^{2} \times \frac{1}{ {x}^{2} } =( 27) {}^{2}$

$\rm \: {x}^{4} + \frac{1}{ x{}^{4} } + 2 = 729$

$\rm \: {x}^{4} + \frac{1}{ {x}^{4} } = 729 - 2$

$\boxed{ \rm \: {x}^{4} + \frac{1}{ {x}^{4} } = 727}$