Question

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

• A. $12$
• B. $9$
• C. $18$
• D. $16$

Answer 1

The correct option is C $18$

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

Squaring both sides,

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

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

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

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

$=>a^2+\frac{1}{a^2}=18$