Question

If the value of $a+\frac{1}{a}=3$, then the value of $a^4+\frac{1}{a^4}$ is equal to.

• A. 45
• B. 47
• C. 49
• D. 51

Answer 1

The correct option is B 47

It is given that $a+\frac{1}{a}=3$
Using identity,
$(a+b{)}^2=a^2+2ab+b^2$,
we have
$a^2+\frac{1}{a^2}=(a+\frac{1}{a}{)}^2-2$
$\Rightarrow$ $a^2+\frac{1}{a^2}=3^2-2$
$\Rightarrow$ $a^2+\frac{1}{a^2}=9-2$
$\Rightarrow$ $a^2+\frac{1}{a^2}=7$

Squaring both sides we get,

$(a^2+\frac{1}{a^2}{)}^2=7^2$
$\Rightarrow$ $(a^2{)}^2+(\frac{1}{a^2}{)}^2+2\left(a^2\right)\left(\frac{1}{a^2}\right)=7^2$
$\Rightarrow$ $a^4+\frac{1}{a^4}+2=49$
$\Rightarrow$ $a^4+\frac{1}{a^4}=47$