Question

If $x=3-\sqrt{8},\mathrm{then}{\left(x-\frac{1}{x}\right)}^2=\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

$$\begin{gathered} \mathrm{Given}:x=3-\sqrt{8} \\ \mathrm{Now},\frac{1}{x}=\frac{1}{3-\sqrt{8}} \\ \mathrm{Multiply}\mathrm{and}\mathrm{divide}\mathrm{RHS}\mathrm{by}\left(3+\sqrt{8}\right),\mathrm{we}\mathrm{get} \\ \Rightarrow \frac{1}{x}=\frac{1}{3-\sqrt{8}}\times \frac{3+\sqrt{8}}{3+\sqrt{8}} \\ \Rightarrow \frac{1}{x}=\frac{3+\sqrt{8}}{{\left(3\right)}^2-{\left(\sqrt{8}\right)}^2}\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow \frac{1}{x}=\frac{3+\sqrt{8}}{9-8} \\ \Rightarrow \frac{1}{x}=\frac{3+\sqrt{8}}{1} \\ \Rightarrow \frac{1}{x}=3+\sqrt{8} \\ \mathrm{Thus}, \\ {\left(x-\frac{1}{x}\right)}^2={\left(\left(3-\sqrt{8}\right)-\left(3+\sqrt{8}\right)\right)}^2 \\ ={\left(3-\sqrt{8}-3-\sqrt{8}\right)}^2 \\ ={\left(-2\sqrt{8}\right)}^2 \\ =32 \end{gathered}$$


Hence, if $x=3-\sqrt{8},\mathrm{then}{\left(x-\frac{1}{x}\right)}^2=\underline{32}.$