Question

If $a+\frac{1}{a}=4$ and $a\ne 0$, find $a^2+\frac{1}{a^2}$.

• A. $14$
• B. $12$
• C. $17$
• D. $19$

Answer 1

The correct option is A $14$

$a+\frac{1}{a}=4$

Squaring both sides,
$=>{(a+\frac{1}{a})}^2=4^2$

$=>a^2+{\left(\frac{1}{a}\right)}^2+2\left(a\right)\left(\frac{1}{a}\right)=16$ ............... $(a+b{)}^2=a^2+2ab+b^2$

$=>a^2+\frac{1}{a^2}+2=16$

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