Question

If $x+\frac{1}{x}=12$, then the value of $x-\frac{1}{x}$ is ________.

• A. $\sqrt{140}$
• B. $\sqrt{120}$
• C. $10$
• D. $11$

Answer 1

The correct option is A $\sqrt{140}$

Given that $x+\frac{1}{x}=12$

Squaring both sides give,

$x^2+\frac{1}{x^2}+2=144$

$x^2+\frac{1}{x^2}=142$ .........(1)

Now ${[x-\frac{1}{x}]}^2=x^2+\frac{1}{x^2}-2$

${[x-\frac{1}{x}]}^2=142-2=140$

$[x-\frac{1}{x}]=\pm \sqrt{140}$