Question

If $2a-\frac{1}{2a}=3$, then $16a^4+\frac{1}{16a^4}$ is equal to:

• A. $1$
• B. $119$
• C. $117$
• D. $121$

Answer 1

The correct option is B $119$

As given$\Rightarrow 2a-\frac{1}{2a}=3$

$\Rightarrow {(2a-\frac{1}{2a})}^2={\left(3\right)}^2$

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

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

$\Rightarrow {(4a^2+\frac{1}{4a^2})}^2={\left(11\right)}^2$

$\Rightarrow 16a^4+\frac{1}{16a^4}+2=121$

$\Rightarrow 16a^4+\frac{1}{16a^4}=121-2$

$\Rightarrow 16a^4+\frac{1}{16a^4}=119$