Find the value of $x^2+\frac{1}{x^2}$ if $x+\frac{1}{x}=3$.

Given:

$x+\frac{1}{x}=3$.

To do:

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

Solution:

We know that,

$(a+b)^2=a^2+2ab+b^2$ $x+\frac{1}{x}=3$

Squaring on both sides, we get,

$(x+\frac{1}{x})^2=(3)^2$

$x^2+2(x)(\frac{1}{x})+\frac{1}{x^2}=9$

$x^2+2+\frac{1}{x^2}=9$

$x^2+\frac{1}{x^2}=9-2$

$x^2+\frac{1}{x^2}=7$.

The value of $x^2+\frac{1}{x^2}$ is $7$.

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Updated on: 10-Oct-2022

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