Question

If $a\ne 0$ and $a-\frac{1}{a}=3$, then find $a^2+\frac{1}{a^2}$.

• A. $11$
• B. $13$
• C. $17$
• D. $19$

Answer 1

The correct option is A $11$

$a-\frac{1}{a}=3$

Squaring both sides,

$=>(a-\frac{1}{a}{)}^2=3^2$

$=>a^2+(\frac{1}{a}{)}^2-2(a\left)\right(\frac{1}{a})=9$

$=>a^2+\frac{1}{a^2}-2=9$

$=>a^2+\frac{1}{a^2}=11$