Question

If $a+\frac{1}{a}=4$ and $a\ne 0$, find $a^4+\frac{1}{a^4}$.

• A. $102$
• B. $194$
• C. $176$
• D. $134$

Answer 1

Answer: (B)

$194$

$a+\frac{1}{a}=4$
Squaring both sides,
$=>(a+\frac{1}{a}{)}^2=4^2$
$=>a^2+(\frac{1}{a}{)}^2+2(a\left)\right(\frac{1}{a})=16$
$=>a^2+\frac{1}{a^2}+2=16$
$=>a^2+\frac{1}{a^2}=14$
Squaring both sides,
$=>(a^2+\frac{1}{a^2}{)}^2={14}^2$
$=>(a^2{)}^2+2(a^2\left)\right(\frac{1}{a^2})+(\frac{1}{a^2}{)}^2=196$
$=>a^4+\frac{1}{a^4}=196-2$
$=>a^4+\frac{1}{a^4}=194$