Question

if x-1/X=3 find the value of x^2+1/x^2 and x=^4+1/x^4​

Answer 1

Correct Question:-

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

★Solution:-

$\rm x- \dfrac{1}{x}=3$

Squaring it will give,

$\rm\implies \Bigg ( x- \dfrac{1}{x} \Bigg )^2 =(x)^2+\Bigg ( \dfrac{1}{x}\Bigg )^2-2 \times x \times \dfrac{1}{x} \qquad\qquad ...[ since, \ (a-b)^2=a^2+b^2-2ab]$

Solving it,

$\rm\implies (3)^2 =x^2+\dfrac{1}{x^2} -2$

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

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

$\rm \implies 11 =x^2+ \dfrac{1}{x^2}$

$\qquad\qquad\qquad\bigstar\boxed{\bf x^2+ \dfrac{1}{x^2}=11}\bigstar$

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Now we will find $\rm x^4 + \dfrac{1}{x^4}$

We know that,

$\boxed{\rm (a^2+b^2)^2 = a^4+b^4+ 2a^2b^2}$

Using this identity,

$\rm\implies\Bigg ( x^2+\dfrac{1}{x^2} \Bigg )^2 = x^4 + \dfrac{1}{x^4} + 2 \times x^2 \times \dfrac{1}{x^2}$

$\rm\implies (11 )^2 = x^4 + \dfrac{1}{x^4} + 2$

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

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

$\rm\implies 119= x^4 + \dfrac{1}{x^4}$

$\qquad\qquad\qquad\bigstar\boxed{\bf x^4+ \dfrac{1}{x^4}=119}\bigstar$

Answer 2

Answer:

$x - \frac{1}{x} = 3$

now,

${x}^{2} + \frac{1}{ {x}^{2} } = (x - \frac{1}{x}) {}^{2} + 2 \times x \times \frac{1}{x} = {3}^{2} + 2 = 9 + 2 = 11 \: \: ans.$

now,

${x}^{4} + \frac{1}{ {x}^{4} } = ( {x}^{2} + \frac{1}{ {x}^{2} } ) {}^{2} - 2 {x}^{2} \times \frac{1}{ {x}^{2} } = {11}^{2} - 2 = 121 - 2 = 119$

Thanks.