Question

Solve the equation:
If $x+\frac{1}{x}=\sqrt{5}$, find $x^4+\frac{1}{4^4}$

Answer 1

Solution:-

$x+\frac{1}{x}=\sqrt{5}\left(\text{Given}\right)$

Squaring both sides, we have

${(x+\frac{1}{x})}^2={\left(\sqrt{5}\right)}^2$

$x^2+\frac{1}{x^2}+2.x.\frac{1}{x}=5(\because {(a+b)}^2=a^2+b^2+2ab)$

$x^2+\frac{1}{x^2}+2=5$

$x^2+\frac{1}{x^2}=5-2$

$x^2+\frac{1}{x^2}=3$

Again squaring both sides, we have

${(x^2+\frac{1}{x^2})}^2={\left(3\right)}^2$

$x^4+\frac{1}{x^4}+2.x^2\frac{1}{x^2}=9$

$x^4+\frac{1}{x^4}+2=9$

$x^4+\frac{1}{x^4}=9-2$

$x^4+\frac{1}{x^4}=7$

Hence the value of $x^4+\frac{1}{x^4}$ is $7$.