Question

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

• A. $9$
• B. $17$
• C. $14$
• D. $8$

Answer 1

The correct option is C $14$

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

Squaring both sides,

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

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

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

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