Question

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

• A. $11$
• B. $21$
• C. $18$
• D. $14$

Answer 1

The correct option is C $18$

$a-\frac{1}{a}=4$
Squaring both sides,
$=>(a-\frac{1}{a}{)}^2=4^2$
$=>a^2+(\frac{1}{a}{)}^2-2(a\left)\right(\frac{1}{a})=16$
$=>a^2+\frac{1}{a^2}-2=16$
$=>a^2+\frac{1}{a^2}=16+2$
$=>a^2+\frac{1}{a^2}=18$