Question

If $(x-\frac{1}{x})=4$ , then the value of $(x^2+\frac{1}{x^2})$ is :

• A. 16
• B. 18
• C. 14
• D. none of these

Answer 1

The correct option is B 18

Given $x-\frac{1}{x}=4$

Squaring on both sides

$(x-\frac{1}{x}{)}^2=4^2$

$x^2+\frac{1}{x^2}-2\left(x\right)\left(\frac{1}{x}\right)=16$

$⟹x^2+\frac{1}{x^2}-2=16$

$⟹x^2+\frac{1}{x^2}=18$