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