Simplify expression $(x\sqrt{y}- y\sqrt{x})/(x\sqrt{y} + y\sqrt{x})$

Question

I'm stuck at the expression: $\displaystyle \frac{x\sqrt{y} -y\sqrt{x}}{x\sqrt{y} + y\sqrt{x}}$.
I need to simplify the expression (by making the denominators rational) and this is what I did:

$$(x\sqrt{y} - y\sqrt{x}) \times (x\sqrt{y} - y\sqrt{x}) = (\sqrt{y} - \sqrt{x})^2$$ Divided by
$$(x\sqrt{y} + y\sqrt{x}) \times (x\sqrt{y} - y\sqrt{x} ) = (x\sqrt{y})^2$$

So I'm left with $\displaystyle \frac{(\sqrt{y} - \sqrt{x})^2}{(x\sqrt{y})^2}$.

This answer is incorrect. Can anyone help me understand what I did wrong? If there is a different approach to solve this it will also be much appreciated. Please explain in steps.

Answer 1

I am assuming your ambiguous notation begins with the task of simplifying:

$$\frac{x\sqrt y - y\sqrt x}{x\sqrt y + y\sqrt x}.$$

Assuming I'm correct, then we can rationalize the denominator (get rid of the factors with square roots), as follows:

Multiply the numerator and denominator by $(x\sqrt{y}-y\sqrt{x})$ to get a difference of squares. Recall that $$(a+b)(a-b) = a^2 - b^2.$$ If you carry out this multiplication, you'll have $$\dfrac{(x\sqrt{y}-y\sqrt{x})^2}{x^2y-xy^2}= \dfrac{x^2y - 2xy\sqrt{xy} + xy^2}{x^2y-xy^2}\; =\; \frac{xy(x-2\sqrt{xy} + y)}{xy(x-y)}\;= \; \frac{x-2\sqrt{xy} + y}{x-y}$$

You seemed to have the right idea, looking at your strategy, to multiply numerator and denominator by $x\sqrt y - y\sqrt x$, but you miscalculated.

Answer 2

Given the question, we can write

\begin{eqnarray} \frac{ x\sqrt{y} - y\sqrt{x} }{ x\sqrt{y} + y\sqrt{x} } &=& \frac{ 1 - \sqrt{y/x} }{ 1 + \sqrt{y/x} }\\ &=& \frac{ \Big( 1 - \sqrt{y/x} \Big)^2 } { \Big( 1 + \sqrt{y/x} \Big) \Big( 1 - \sqrt{y/x} \Big)}\\ &=& \frac{ 1 + y/x - 2\sqrt{y/x} }{ 1 - y/x }\\ &=& \frac{ x + y - 2\sqrt{xy} }{ x - y }. \end{eqnarray}

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Or

\begin{eqnarray} \frac{ x\sqrt{y} - y\sqrt{x} }{ x\sqrt{y} + y\sqrt{x} } &=& \frac{ \sqrt{x} - \sqrt{y} }{ \sqrt{x} + \sqrt{y} }\\ &=& \frac{ \Big( \sqrt{x} - \sqrt{y} \Big)^2 } { \Big( \sqrt{x} + \sqrt{y} \Big) \Big( \sqrt{x} - \sqrt{y} \Big)}\\ &=& \frac{ x + y - 2\sqrt{xy} }{ x - y }. \end{eqnarray}

Answer 3

The answer can be simplified even further (using the difference of two squares):

$$\frac{x-2\sqrt{xy} + y}{x-y} = \frac{(\sqrt{x} - \sqrt{y})^2}{(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})} = \frac{1}{\sqrt{x} - \sqrt{y}} \text{ or } \frac{\sqrt{x} - \sqrt{y}}{x+y}$$