Question

If $\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{6},$ then the value of $\left(x^4-190+\frac{1}{x^4}\right)$ is

Answer 1

$\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{6}$

Squaring both sides, $x+\frac{1}{x}+2=6$

$\Rightarrow (x+\frac{1}{x})=4$

Now, $x^4+\frac{1}{x^4}=190=(x^2+\frac{1}{x^2}{)}^2-2-190$

$=\left(\right(x+\frac{1}{x}{)}^2+2{)}^2-192$

$=\left(\right(4{)}^2+2{)}^2-192$

$=(18{)}^2-192$

$=232$