Question

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

• A. $18$
• B. $11$
• C. $3$
• D. $9$

Answer 1

Answer: (B)

$11$

$a-\frac{1}{a}=3$
Squaring both sides,
$=>(a-\frac{1}{a}{)}^2=3^2$
$=>a^2+(\frac{1}{a}{)}^2-2(a\left)\right(\frac{1}{a})=9$
$=>a^2+\frac{1}{a^2}-2=9$
$=>a^2+\frac{1}{a^2}=11$