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Question:

If $x+\frac{1}{x}=\sqrt{5}$, find the value of $x^{2}+\frac{1}{x^{2}}$ and $x^{4}+\frac{1}{x^{4}}$

Solution:

We have,

$(x+1 / x)^{2}=x^{2}+(1 / x)^{2}+2 * x * 1 / x$

$\Rightarrow(x+1 / x)^{2}=x^{2}+1 / x^{2}+2$

$\Rightarrow(\sqrt{5})^{2}=x^{2}+\frac{1}{x^{2}}+2\left[\therefore x+\frac{1}{x}=\sqrt{5}\right]$

$\Rightarrow 5=x^{2}+1 / x^{2}+2$

$\Rightarrow x^{2}+1 / x^{2}=3 \ldots(1)$

Now, $\left(x^{2}+1 / x^{2}\right)^{2}=x^{4}+1 / x^{4}+2 * x^{2} * 1 / x^{2}$

$\Rightarrow\left(x^{2}+1 \times 2\right)^{2}=x^{4}+1 / x^{4}+2$

$\Rightarrow 9=x^{4}+1 / x^{4}+2\left[\therefore x^{2}+1 / x^{2}=3\right]$

$\Rightarrow x^{4}+1 / x^{4}=7$

Hence, $x^{2}+1 / x^{2}=3 ; x^{4}+1 / x^{4}=7 .$

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