Question

if (x-1/x=√3 then find the value of (x^2+1/x^2)and(x^4+1/x^4)​

Answer 1

★Given:-

$\sf x - \dfrac{1}{x} = \sqrt{3}$

★To find:-

• $\sf x^2 + \dfrac{1}{x^2}$
• $\sf x^4 + \dfrac{1}{x^4}$

★Solution:-

We know that $(a-b)^2 = a^2 +b^2 -2ab$

Using this identity,

$\sf \Bigg ( x - \dfrac{1}{x} \Bigg )^2 = x^2 + \dfrac{1}{x^2} - 2 \times x \times \dfrac{1}{x}$

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

$\sf \implies 3= x^2 + \dfrac{1}{x^2} - 2$

$\sf \implies 3+2= x^2 + \dfrac{1}{x^2}$

$\sf \implies 5= x^2 + \dfrac{1}{x^2}$

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

Squaring $x^2 + \dfrac{1}{x^2}$ and using identity (a+b)² = a²+b² + 2ab

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

$\sf \implies (5 )^2 = x^4 + \dfrac{1}{x^4} + 2$

$\sf \implies 25 = x^4 + \dfrac{1}{x^4} + 2$

$\sf \implies 25 - 2= x^4 + \dfrac{1}{x^4}$

$\sf \implies 23= x^4 + \dfrac{1}{x^4}$

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Value of $\sf x^2 + \dfrac{1}{x^2} = 5$

And value of $\sf x^4 + \dfrac{1}{x^4} = 23$

Hope it helps.