How to solve the equation

Question

If $x+y = 5$, $xy = 1$ and $x > y$, then $\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}= ?$ The answer key gives for the asnwer $\frac{\sqrt{21}}{3}$, $\frac{7}{\sqrt{21}}$, $\frac{\sqrt{7}}{\sqrt{3}}$, $\frac{7}{3}$. $$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} * \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}= \frac{x+2\sqrt{x}\sqrt{y}+y}{x-y}= \frac{5+2\sqrt{1}}{x-y}= \frac{7}{x-y}$$ Once $$x=5-y $$ $$(5-y)y=1 $$ $$y^2-5y+1 $$ $$y=\frac{5\pm \:\sqrt{21}}{2} $$ Considering $x=5-y$, in order to $x>y$, the value of y should be: $y=\frac{5-\sqrt{21}}{2} $. Therefore, x will be $y=\frac{5+\sqrt{21}}{2}.$ Now solving $x-y:$

$$x-y\Rightarrow \frac{5+\sqrt{21}}{2}- \frac{5-\sqrt{21}}{2}$$ $$x-y\Rightarrow \frac{5-5+\sqrt{21}+\sqrt{21}}{2}$$ $$x-y\Rightarrow \frac{2\sqrt{21}}{2}$$ $$x-y= \sqrt{21}$$ Back to $\frac{7}{x-y}$ $$\frac{7}{x-y}= \frac{7}{\sqrt{21}} $$ $$∴\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}= \frac{7}{\sqrt{21}} $$

I did solve it but it wasn't a nice solution. Can anyone give a cleaver and more interesting solution? I am studying for a test and wish there were a quick way to solve this kind of question, I would appreciate some interesting ideas.

Answer 1

What about multiplying it by $\sqrt{x} - \sqrt{y}$ and compare it with your result?

\begin{align*} \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\times\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x} - \sqrt{y}} = \frac{x-y}{x - 2\sqrt{xy} + y} = \frac{x-y}{3} = \frac{7}{x-y} \end{align*}

Consequently, $(x-y)^{2} = 21$. Since $x > y$, one has that $x - y = \sqrt{21}$.

Can you take it from here?

Answer 2

Hint

If $z=\dfrac{\sqrt x+\sqrt y}{\sqrt x-\sqrt y},$

$$z^2=\dfrac{x+y+2\sqrt{xy}}{?}=\dfrac{5+2}{5-2}$$

as $\sqrt{xy}=1$

Now $\sqrt x\pm\sqrt y>0\implies z>0$