Question

If $x=\frac{2}{\sqrt{10}-\sqrt{8}}\mathrm{and}y=\frac{2}{\sqrt{10}+2\sqrt{2}},\mathrm{then}{(x-y)}^2=\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ x=\frac{2}{\sqrt{10}-\sqrt{8}} \\ y=\frac{2}{\sqrt{10}+2\sqrt{2}} \\ \mathrm{Now}, \\ x=\frac{2}{\sqrt{10}-\sqrt{8}} \\ \mathrm{Multiply}\mathrm{and}\mathrm{divide}\mathrm{RHS}\mathrm{by}\left(\sqrt{10}+\sqrt{8}\right),\mathrm{we}\mathrm{get} \\ \Rightarrow x=\frac{2}{\sqrt{10}-\sqrt{8}}\times \frac{\sqrt{10}+\sqrt{8}}{\sqrt{10}+\sqrt{8}} \\ \Rightarrow x=\frac{2\left(\sqrt{10}+\sqrt{8}\right)}{{\left(\sqrt{10}\right)}^2-{\left(\sqrt{8}\right)}^2}\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow x=\frac{2\left(\sqrt{10}+\sqrt{8}\right)}{10-8} \\ \Rightarrow x=\frac{2\left(\sqrt{10}+\sqrt{8}\right)}{2} \\ \Rightarrow x=\frac{\left(\sqrt{10}+\sqrt{8}\right)}{1} \\ \Rightarrow x=\sqrt{10}+\sqrt{8} \\ \Rightarrow x=\sqrt{10}+2\sqrt{2} \\ y=\frac{2}{\sqrt{10}+2\sqrt{2}} \\ \mathrm{Multiply}\mathrm{and}\mathrm{divide}\mathrm{RHS}\mathrm{by}\left(\sqrt{10}-2\sqrt{2}\right),\mathrm{we}\mathrm{get} \\ \Rightarrow y=\frac{2}{\sqrt{10}+2\sqrt{2}}\times \frac{\sqrt{10}-2\sqrt{2}}{\sqrt{10}-2\sqrt{2}} \\ \Rightarrow y=\frac{2\left(\sqrt{10}-2\sqrt{2}\right)}{{\left(\sqrt{10}\right)}^2-{\left(2\sqrt{2}\right)}^2}\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow y=\frac{2\left(\sqrt{10}-2\sqrt{2}\right)}{10-8} \\ \Rightarrow y=\frac{2\left(\sqrt{10}-2\sqrt{2}\right)}{2} \\ \Rightarrow y=\frac{\left(\sqrt{10}-2\sqrt{2}\right)}{1} \\ \Rightarrow y=\sqrt{10}-2\sqrt{2} \\ \mathrm{Thus}, \\ {\left(x-y\right)}^2={\left(\left(\sqrt{10}+2\sqrt{2}\right)-\left(\sqrt{10}-2\sqrt{2}\right)\right)}^2 \\ ={\left(\sqrt{10}+2\sqrt{2}-\sqrt{10}+2\sqrt{2}\right)}^2 \\ ={\left(4\sqrt{2}\right)}^2 \\ =32 \end{gathered}$$


Hence, if $x=\frac{2}{\sqrt{10}-\sqrt{8}}\mathrm{and}y=\frac{2}{\sqrt{10}+2\sqrt{2}},\mathrm{then}{(x-y)}^2=\underline{32}.$