Question

If $x+\frac{1}{x}=12$, then find the value of $(x^2+\frac{1}{x^2})$.

Answer 1

Given, $x+\frac{1}{x}=12$.

We know, $(a+b{)}^2=a^2+2ab+b^2$.

Then,

${\left(x+\frac{1}{x}\right)}^2=144$

$\Rightarrow x^2+\frac{1}{x^2}+2\times x\times \frac{1}{x}=144$

$\Rightarrow x^2+\frac{1}{x^2}+2=144$

$\Rightarrow x^2+\frac{1}{x^2}=144-2$

$\Rightarrow x^2+\frac{1}{x^2}=142$.

Therefore, the required answer is $142$.