If $ x=3+\sqrt{8} $, find the value of $ x^{2}+\frac{1}{x^{2}} $.

Given:

\( x=3+\sqrt{8} \)

To do: 

We have to find the value of \( x^{2}+\frac{1}{x^{2}} \).

Solution:

We know that,

Rationalising factor of a fraction with denominator ${\sqrt{a}}$ is ${\sqrt{a}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}-\sqrt{b}}$ is ${\sqrt{a}+\sqrt{b}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}+\sqrt{b}}$ is ${\sqrt{a}-\sqrt{b}}$.

Therefore,

$x=3+\sqrt{8}$

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

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

$=\frac{3-\sqrt{8}}{(3)^{2}-(\sqrt{8})^{2}}$

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

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

$=3-\sqrt{8}$

Therefore,

$x+\frac{1}{x}=3+\sqrt{8}+3-\sqrt{8}=6$

Squaring both sides, we get,

$(x+\frac{1}{x})^{2}=(6)^{2}$

$\Rightarrow x^{2}+\frac{1}{x^{2}}+2 \times x \times \frac{1}{x}=36$

$\Rightarrow x^{2}+\frac{1}{x^{2}}+2=36$

$\Rightarrow x^{2}+\frac{1}{x^{2}}=36-2$

$\Rightarrow x^{2}+\frac{1}{x^{2}}=34$

The value of \( x^{2}+\frac{1}{x^{2}} \) is $34$. 

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Updated on: 10-Oct-2022

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