Question

If $2x-\frac{1}{2x}=3$, find the value of $16x^4+\frac{1}{16x^4}$

• A. $111$
• B. $117$
• C. $119$
• D. $121$

Answer 1

The correct option is C $119$

$(2x-\frac{1}{2x}{)}^2=4x^2+\frac{1}{4x^2}-2=3^2=9$

$4x^2+\frac{1}{4x^2}=11$

squaring both sides

$(4x^2+\frac{1}{4x^2}{)}^2=16x^4+\frac{1}{16x^4}+2=121$

so $16x^4+\frac{1}{16x^4}=119$