Question

If $2x-\frac{1}{2x}=4$, then find $4x^2+\frac{1}{4x^2}$.

• A. $16$
• B. $20$
• C. $18$
• D. $24$

Answer 1

The correct option is C $18$

Given, $2x-\frac{1}{2x}=4$
Squaring both sides,
${(2x-\frac{1}{2x})}^2=16$
Applying the formula ${(a-b)}^2=a^2+b^2-2ab$
$(2x{)}^2+{\left(\frac{1}{2x}\right)}^2-2\times 2x\times \frac{1}{2x}=16$
$4x^2+\frac{1}{4x^2}-2=16$
$\Rightarrow 4x^2+\frac{1}{4x^2}=18$