Question

If a² – 2a – 1 = 0, then $a^2+\frac{1}{a^2}$ = __________.

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ {a}^2-2a-1=0 \\ \Rightarrow a^2-1=2a \\ \mathrm{Dividing}\text{'}a\text{'}\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow \frac{a^2-1}{a}=\frac{2a}{a} \\ \Rightarrow \frac{a^2}{a}-\frac{1}{a}=2 \\ \Rightarrow a-\frac{1}{a}=2 \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow {\left(a-\frac{1}{a}\right)}^2=2^2 \\ \Rightarrow a^2+{\left(\frac{1}{a}\right)}^2-2a\frac{1}{a}=4\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{\left(a-b\right)}^2=a^2+b^2-2ab\right) \\ \Rightarrow a^2+\frac{1}{a^2}-2=4 \\ \Rightarrow a^2+\frac{1}{a^2}=6 \end{gathered}$$

Hence, if a² – 2a – 1 = 0, then $a^2+\frac{1}{a^2}$ = 6.