Question

If $2a+\frac{1}{2a}=3$, then $16a^4+\frac{1}{16a^4}=?$

• A. $11$
• B. $119$
• C. $117$
• D. $121$

Answer 1

Answer: (B)

Given:

$2a+\frac{1}{2a}=3$

Squaring on both sides

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

$\Rightarrow 4a^2+\frac{1}{4a^2}-2\left(2a\right)\left(\frac{1}{2a}\right)=9$ [Since, $(a-b{)}^2=a^2+b^2-2ab)$]

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

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

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

Again squaring on both sides,

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

$\Rightarrow 16a^4+\frac{1}{16a^4}+2\left(4a^2\right)\left(\frac{1}{4a^2}\right)=121$

[Since $(a+b{)}^2=a^2+b^2+2ab)$]

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

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

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

Hence, Option A is correct.