Question

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

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ a+\frac{1}{a}=-2 \\ \mathrm{Now}, \\ a+\frac{1}{a}=-2 \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow {\left(a+\frac{1}{a}\right)}^2={\left(-2\right)}^2 \\ \Rightarrow a^2+{\left(\frac{1}{a}\right)}^2+2a\frac{1}{a}=4\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{\left(a+b\right)}^2=a^2+b^2+2ab\right) \\ \Rightarrow a^2+\frac{1}{a^2}+2=4 \\ \Rightarrow a^2+\frac{1}{a^2}=4-2 \\ \Rightarrow a^2+\frac{1}{a^2}=2 \end{gathered}$$

Hence, if $a+\frac{1}{a}=-2,\mathrm{then}{a}^2+\frac{1}{a^2}=\underline{2}.$