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