Question

If $x+\frac{1}{x}=12,$ find the value of $x-\frac{1}{x}=12,$

Answer 1

$x+\frac{1}{x}=12$, $x-\frac{1}{x}$

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

$(12{)}^2-2=x^2+\frac{1}{x^2}=144-2$

$=142$

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

$=142-2$

$=140$

$x-\frac{1}{x}=\sqrt{140}$

$=2\sqrt{35}$