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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