Question

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

• A. $190$
• B. $194$
• C. $192$
• D. $193$

Answer 1

The correct option is 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,

$=>(a^2+\frac{1}{a^2}{)}^2={14}^2$

$=>(a^2{)}^2+2(a^2\left)\right(\frac{1}{a^2})+(\frac{1}{a^2}{)}^2=196$

$=>a^4+\frac{1}{a^4}=196-2$

$=>a^4+\frac{1}{a^4}=194$