Question

If $x=\frac{2}{\sqrt{3}-\sqrt{5}}\mathrm{and}y=\frac{2}{\sqrt{3}+\sqrt{5}}$, then x + y = __________.

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ x=\frac{2}{\sqrt{3}-\sqrt{5}} \\ y=\frac{2}{\sqrt{3}+\sqrt{5}} \\ \mathrm{Now}, \\ x=\frac{2}{\sqrt{3}-\sqrt{5}} \\ \mathrm{Multiply}\mathrm{and}\mathrm{divide}\mathrm{RHS}\mathrm{by}\left(\sqrt{3}+\sqrt{5}\right),\mathrm{we}\mathrm{get} \\ \Rightarrow x=\frac{2}{\sqrt{3}-\sqrt{5}}\times \frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}+\sqrt{5}} \\ \Rightarrow x=\frac{2\left(\sqrt{3}+\sqrt{5}\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{5}\right)}^2}\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow x=\frac{2\left(\sqrt{3}+\sqrt{5}\right)}{3-5} \\ \Rightarrow x=\frac{2\left(\sqrt{3}+\sqrt{5}\right)}{-2} \\ \Rightarrow x=\frac{-\left(\sqrt{3}+\sqrt{5}\right)}{1} \\ \Rightarrow x=-\sqrt{3}-\sqrt{5} \\ y=\frac{2}{\sqrt{3}+\sqrt{5}} \\ \mathrm{Multiply}\mathrm{and}\mathrm{divide}\mathrm{RHS}\mathrm{by}\left(\sqrt{3}-\sqrt{5}\right),\mathrm{we}\mathrm{get} \\ \Rightarrow y=\frac{2}{\sqrt{3}+\sqrt{5}}\times \frac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}} \\ \Rightarrow y=\frac{2\left(\sqrt{3}-\sqrt{5}\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{5}\right)}^2}\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow y=\frac{2\left(\sqrt{3}-\sqrt{5}\right)}{3-5} \\ \Rightarrow y=\frac{2\left(\sqrt{3}-\sqrt{5}\right)}{-2} \\ \Rightarrow y=\frac{-\left(\sqrt{3}-\sqrt{5}\right)}{1} \\ \Rightarrow y=-\sqrt{3}+\sqrt{5} \\ \mathrm{Thus}, \\ x+y=\left(-\sqrt{3}-\sqrt{5}\right)+\left(-\sqrt{3}+\sqrt{5}\right) \\ =\left(-\sqrt{3}-\sqrt{5}-\sqrt{3}+\sqrt{5}\right) \\ =\left(-2\sqrt{3}\right) \\ =-2\sqrt{3} \end{gathered}$$


Hence, x + y = $\underline{-2\sqrt{3}}$.