Question

If $2x-\frac{1}{2x}=13$, find the values of $16x^4+\frac{1}{16x^4}$

Answer 1

We have,

If $2x-\frac{1}{2x}=13$

Then, squaring both side and we get,

${(2x-\frac{1}{2x})}^2={\left(13\right)}^2$

$4x^2+\frac{1}{4x^2}-2\times 2x\times \frac{1}{2x}=169$

$4x^2+\frac{1}{4x^2}=169+2$

$4x^2+\frac{1}{4x^2}=171$

$4x^2+\frac{1}{4x^2}=171$

Again, squaring both side and we get,

${(4x^2+\frac{1}{4x^2})}^2={\left(171\right)}^2$

$16x^4+\frac{1}{16x^4}+2\times 4x^2\times \frac{1}{4x^2}=29241$

$16x^4+\frac{1}{16x^4}=29241-2$

$16x^4+\frac{1}{16x^4}=29239$

Hence, this is the answer.