If $x+\frac{1}{x}=4$, then find, $x^{2}+\frac{1}{x^2}=?$

Given: $x+\frac{1}{x}=4$.

To do: To find $x^{2}+\frac{1}{x^2}$

Solution:

Given, $x+\frac{1}{x}=4$

On squaring both sides,

( x+\frac{1}{x})^2=4^2

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

times x\times\frac{1}{x}=16$

$\Rightarrow x^2+\frac{1}{x^2}+2=16$

$\Rightarrow x^2+\frac{1}{x^2}=16-2$

$\Rightarrow x^2+\frac{1}{x^2}=14$

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Updated on: 10-Oct-2022

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