Solve the following

Let us consider the following equation:

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

Squaring both sides, we get:

$\left(x+\frac{1}{x}\right)^{2}=(20)^{2}=400$

$\Rightarrow\left(x+\frac{1}{x}\right)^{2}=400$

$\Rightarrow x^{2}+2 \times x \times \frac{1}{x}+\left(\frac{1}{x}\right)^{2}=400 \quad\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

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

$\Rightarrow x^{2}+\frac{1}{x^{2}}=398$      (Subtracting 2 from both sides)

Thus, the answer is 398.