Question

If $x=9-4\sqrt{5}$, find the value of $x^2+\frac{1}{x^2}$.

Answer 1

$$\begin{gathered} x=9-4\sqrt{5}.....\left(1\right) \\ \Rightarrow \frac{1}{x}=\frac{1}{9-4\sqrt{5}} \\ \Rightarrow \frac{1}{x}=\frac{1}{9-4\sqrt{5}}\times \frac{9+4\sqrt{5}}{9+4\sqrt{5}} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{x}=\frac{9+4\sqrt{5}}{9^2-{\left(4\sqrt{5}\right)}^2} \\ \Rightarrow \frac{1}{x}=\frac{9+4\sqrt{5}}{81-80} \\ \Rightarrow \frac{1}{x}=9+4\sqrt{5}.....\left(2\right) \end{gathered}$$
Adding (1) and (2), we get
$$\begin{gathered} x+\frac{1}{x}=9-4\sqrt{5}+9+4\sqrt{5} \\ \Rightarrow x+\frac{1}{x}=18 \end{gathered}$$
Squaring on both sides, we get
$$\begin{gathered} {\left(x+\frac{1}{x}\right)}^2={18}^2 \\ \Rightarrow x^2+\frac{1}{x^2}+2\times x\times \frac{1}{x}=324 \\ \Rightarrow x^2+\frac{1}{x^2}=324-2=322 \end{gathered}$$
Thus, the value of $x^2+\frac{1}{x^2}$ is 322.