Question

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

Answer 1

$x+\frac{1}{x}=20\text{Squaring on both sides},{(x+\frac{1}{x})}^2=(20{)}^2{x}^2+2\times x\times \frac{1}{x}+\frac{1}{x^2}=400\Rightarrow x^2+2+\frac{1}{x^2}=400\Rightarrow x^2+\frac{1}{x^2}=400-2=398\therefore x^2+\frac{1}{x^2}=398$