If $x + \frac{1}{x} =20$, find the value of $x^2 + \frac{1}{x^2}$.

Given:

$x + \frac{1}{x} =20$

To do:

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

Solution:

The given expression is $x + \frac{1}{x} =20$. Here, we have to find the value of $x^2 + \frac{1}{x^2}$. So, by squaring the given expression and using the identity $(a+b)^2=a^2+2ab+b^2$, we can find the value of $x^2 + \frac{1}{x^2}$.

$(a+b)^2=a^2+2ab+b^2$...................(i)

Now,

$x + \frac{1}{x} =20$

Squaring on both sides, we get,

$(x+\frac{1}{x})^2=(20)^2$

$x^2+2\times x \times \frac{1}{x}+(\frac{1}{x})^2=400$           [Using (i)]

$x^2+2+\frac{1}{x^2}=400$

$x^2+\frac{1}{x^2}=400-2$                 (Transposing $2$ to RHS)

$x^2+\frac{1}{x^2}=398$

Hence, the value of $x^2+\frac{1}{x^2}$ is $398$.