Question

If $x=\left(7+5\sqrt{2}\right)$, then $\left(x^2+\frac{1}{x^2}\right)$=?
(a) 160
(b) 198
(c) 189
(d) 156

Answer 1

(b) 198

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ x=7+5\sqrt{2} \\ \mathrm{Now}, \\ \frac{1}{x}=\frac{1}{7+5\sqrt{2}}\times \frac{7-5\sqrt{2}}{7-5\sqrt{2}}=\frac{7-5\sqrt{2}}{49-50}=-\left(7-5\sqrt{2}\right)=5\sqrt{2}-7 \\ \mathrm{And}, \\ x+\frac{1}{x}=7+5\sqrt{2}+5\sqrt{2}-7=10\sqrt{2} \\ \mathrm{W}e\mathrm{know}: \\ {\left(x+\frac{1}{x}\right)}^2=x^2+\frac{1}{x^2}+2 \\ \Rightarrow {\left(10\sqrt{2}\right)}^2=x^2+\frac{1}{x^2}+2 \\ \Rightarrow x^2+\frac{1}{x^2}=200-2 \\ =198 \end{gathered}$$