If x = 3+√8, find the value of

Question:

If $x=3+\sqrt{8}$, find the value of

$\left(x^{2}+\frac{1}{x^{2}}\right)$

Solution:

Given, $x=3+\sqrt{8}$

$\left(x^{2}+\frac{1}{x^{2}}\right)$

We have, $x=3+\sqrt{8}$,

$\frac{1}{x}=\frac{1}{3+\sqrt{8}}$

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

$3-\sqrt{8}$ for $\frac{1}{3+\sqrt{8}}$

$\frac{1}{x}=\frac{3-\sqrt{8}}{(3+\sqrt{8})(3-\sqrt{8})}$

Since, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$\frac{1}{x}=\frac{3-\sqrt{8}}{9-8}$

$\frac{1}{x}=3-\sqrt{8}$

$\left(x^{2}+\frac{1}{x^{2}}\right)=\left((3+\sqrt{8})^{2}(3-\sqrt{8})^{2}\right)$

$\left(x^{2}+\frac{1}{x^{2}}\right)=((9+8+6 \sqrt{8})+(9+8-6 \sqrt{8}))$

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