Question

If $x=\sqrt{6}+\sqrt{5},\mathrm{then}{x}^2+\frac{1}{x^2}-2=\_\_\_\_\_\_\_\_\_.$

Answer 1

$$\begin{gathered} \mathrm{Given}:x=\sqrt{6}+\sqrt{5} \\ \mathrm{Now},\frac{1}{x}=\frac{1}{\sqrt{6}+\sqrt{5}} \\ \mathrm{Multiply}\mathrm{and}\mathrm{divide}\mathrm{RHS}\mathrm{by}\left(\sqrt{6}-\sqrt{5}\right),\mathrm{we}\mathrm{get} \\ \Rightarrow \frac{1}{x}=\frac{1}{\sqrt{6}+\sqrt{5}}\times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}} \\ \Rightarrow \frac{1}{x}=\frac{\sqrt{6}-\sqrt{5}}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{5}\right)}^2}\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow \frac{1}{x}=\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ \Rightarrow \frac{1}{x}=\frac{\sqrt{6}-\sqrt{5}}{1} \\ \Rightarrow \frac{1}{x}=\sqrt{6}-\sqrt{5} \\ \mathrm{Thus}, \\ {x}^2+\frac{1}{x^2}-2={\left(x-\frac{1}{x}\right)}^2 \\ ={\left(\left(\sqrt{6}+\sqrt{5}\right)-\left(\sqrt{6}-\sqrt{5}\right)\right)}^2 \\ ={\left(\sqrt{6}+\sqrt{5}-\sqrt{6}+\sqrt{5}\right)}^2 \\ ={\left(2\sqrt{5}\right)}^2 \\ =20 \end{gathered}$$


Hence, if $x=\sqrt{6}+\sqrt{5},\mathrm{then}{x}^2+\frac{1}{x^2}-2=\underline{20}.$