Question

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

Answer 1

Applying the formula, ${(a-b)}^2=a^2+b^2-2ab$
${(a-\frac{1}{a})}^2=a^2+{\left(\frac{1}{a}\right)}^2-2a\times \frac{1}{a}=a^2+\frac{1}{a^2}-2$
$=>a^2+\frac{1}{a^2}={(a-\frac{1}{a})}^2+2$
Substituting $a-\frac{1}{a}=4$,
$=>a^2+\frac{1}{a^2}=4^2+2=16+2=18$