Question

${(a+\frac{1}{a}+2)}^2=4$, then find $a^2+\frac{1}{a^2}$ give that $(a^2+\frac{1}{a^2})$ cannot be negative.

Answer 1

Consider the given,

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

$(a+\frac{1}{a}+2)=2$

$a+\frac{1}{a}=0$

Taking square both sides,

${(a+\frac{1}{a})}^2=0$

$a^2+\frac{1}{a^2}+2=0$

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