Question

If $a^2+\frac{1}{a^2}=11$, then find the value of $(a-\frac{1}{a})$

Answer 1

Applying the formula, ${(a-b)}^2=a^2+b^2-2ab$
${(a-\frac{1}{a})}^2=a^2+{\left(\frac{1}{a}\right)}^2-2a\times \frac{1}{a}=a^2+\frac{1}{a^2}-2$
So, ${(a-\frac{1}{a})}^2=a^2+\frac{1}{a^2}-2$

Substituting $a^2+{\left(\frac{1}{a}\right)}^2=11$,

$=>{(a-\frac{1}{a})}^2=11-2=9$

$a-\frac{1}{a}=\sqrt{9}=3$