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I've been trying to solve this expression for at least two hours now... And I always get stuck towards the end, I don't know what I'm missing.

$\frac 1{xy} \times (\sqrt{xy} - \frac{xy}{x-\sqrt{xy}})\times (\sqrt{xy} + \frac{xy}{x+\sqrt{xy}})$

My first step was to rationalize the fractions inside the parenthesis like so

$\frac 1{xy} \times (\sqrt{xy} - \frac{xy(x+\sqrt{xy})}{(x-\sqrt{xy})(x+\sqrt{xy})})\times (\sqrt{xy} + \frac{xy(x-\sqrt{xy})}{(x+\sqrt{xy})(x-\sqrt{xy})})$

to get

$\frac 1{xy} \times (\sqrt{xy} - \frac{xy(x+\sqrt{xy})}{(x^2-xy)})\times (\sqrt{xy} + \frac{xy(x-\sqrt{xy})}{(x^2-xy)})$

then

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

and then I'm kind of lost, nothing I've tried works. I tried grouping each fraction by x and simplyfing removing it, and then computing the lcm inside the parenthesis in order to subtract the $\sqrt{xy}$. Or the other way around, first I did the lcm and subtracted and then simplyfied. I even tried multiplying the first factor by the second and simplifying as I went on. I tried using Wolfram Alpha to help me with each step, too. I think my calculations are correct, I'm just missing some simplification or something similar. The result should be

$\frac{x-4y}{x-y}$

I was able to get the $x-y$ but not the $x-4y$. I hope I didn't mess up the equations in the question, I'm really tired.

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    $\begingroup$ Do not rationalize the denominators first: put each parenthesized factor over a common denominator, then expand the product in the numerator (the denominator will be the product of the two conjugates so it will simplify immediately). $\endgroup$ – NickD Nov 5 '18 at 21:31
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We have

$$\frac 1{xy} \times (\sqrt{xy} - \frac{xy}{x-\sqrt{xy}})\times (\sqrt{xy} + \frac{xy}{x+\sqrt{xy}}) =\frac 1{xy} \times \left(xy - \frac{x^2y^2}{x^2-xy}+\frac{xy\sqrt{xy}}{x+\sqrt{xy}}-\frac{xy\sqrt{xy}}{x-\sqrt{xy}}\right)=$$

$$=1 - \frac{xy}{x^2-xy}+\frac{\sqrt{xy}}{x+\sqrt{xy}}-\frac{\sqrt{xy}}{x-\sqrt{xy}}=$$

$$=1 - \frac{xy}{x^2-xy}+\frac{x\sqrt{xy}-xy}{x^2-xy}-\frac{x\sqrt{xy}+xy}{x^2-xy}=$$

$$=\frac{x^2-4xy}{x^2-xy}=\frac{x-4y}{x-y}$$

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  • $\begingroup$ Got it, thank you very much. I didn't think about just multiplying first and then simplify, Or maybe I did and I messed up the calculations. I'm tired I don't even remember haha But from this answer I think I understood the rules, I was just getting lost among all the various operands... Thank you again. $\endgroup$ – Paul Nov 5 '18 at 21:40
  • $\begingroup$ @Paul You are welcome! I'm going to revise your calculation to find the mistake! $\endgroup$ – gimusi Nov 5 '18 at 21:43
  • $\begingroup$ @Paul I can't see mistakes in your derivation, if you proceed multiplying as I did you should obtain the same answer! $\endgroup$ – gimusi Nov 5 '18 at 21:44
  • $\begingroup$ I think I did alright up to that point, then stuff starts to go bad. I have 10 sheets of papers with different attempts and all are just slightly different from each other in the end, so I think I either messed up a sign or just made a writing mistake. I need to rest. Thank you very much again! $\endgroup$ – Paul Nov 5 '18 at 21:45
  • $\begingroup$ @Paul That's a good idea! Tomorrow all things will be clear. Bye :) $\endgroup$ – gimusi Nov 5 '18 at 21:46
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For simplicity, set $z=\sqrt{xy}$, with $z^2=xy$.

Then your expression becomes \begin{align} \frac {1}{xy}\left(z - \frac{xy}{x-z}\right)\left(z + \frac{xy}{x+z}\right) &=\frac {1}{xy}\frac{xz-z^2-xy}{x-z}\frac{xz+z^2+xy}{x+z}\\[4px] &=\frac {1}{xy}\frac{xz-2xy}{x-z}\frac{xz+2xy}{x+z}\\[4px] &=\frac {1}{xy}\frac{x^2(z-2y)(z+2y)}{(x-z)(x+z)}\\[4px] &=\frac {1}{xy}\frac{x^2(z^2-4y^2)}{x^2-z^2}\\[4px] &=\frac {1}{y}\frac{x(xy-4y^2)}{x^2-xy}\\[4px] &=\frac {1}{y}\frac{xy(x-4y)}{x(x-y)}\\[4px] &=\frac{x-4y}{x-y} \end{align}

A similar strategy might be to set $a=\sqrt{x}$ and $b=\sqrt{y}$ (assuming both are positive, but the same holds when both are negative). Then we have $$ \sqrt{xy}-\frac{xy}{x-\sqrt{xy}}=ab-\frac{a^2b^2}{a^2-ab}=ab-\frac{ab^2}{a-b} =ab\left(1-\frac{b}{a-b}\right)=\frac{ab(a-2b)}{a-b} $$ Similarly $$ \sqrt{xy}+\frac{xy}{x+\sqrt{xy}}=\frac{ab(a+2b)}{a+b} $$ so your expression becomes $$ \frac{1}{a^2b^2}\frac{a^2b^2(a-2b)(a+2b)}{(a-b)(a+b)}= \frac{a^2-4b^2}{a^2-b^2}=\frac{x-4y}{x-y} $$

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  • $\begingroup$ I understand what you did, but is there a way to do this without this "trick"? I mean by removing the square roots using the various rules and methods. Also you said z^2=xy, but you still also leave xy in the rest of the problem. Shouldn't it turn into z^2? $\endgroup$ – Paul Nov 5 '18 at 21:28
  • $\begingroup$ @Paul I added an alternative way. Anyway, there is no “trick”, just a way to get rid of the frightening $\sqrt{xy}$. $\endgroup$ – egreg Nov 5 '18 at 21:31
  • $\begingroup$ Even in your alternative way you still "cheated" by turning the square root in a different variable. I'm still grateful because I learned that I can do that, and it looks like a simpler way. :) @gimusi 's anwer is more akin to what I was looking for, just removing the sqrt using sqrt properties (like rationalizing the denominator). $\endgroup$ – Paul Nov 5 '18 at 21:43
  • $\begingroup$ @Paul There is no need to inflict pain to oneself if it's possible to avoid it. That's one of the reasons for the development of algebra. $\endgroup$ – egreg Nov 5 '18 at 21:44
  • $\begingroup$ I agree and thank you for teaching me. $\endgroup$ – Paul Nov 5 '18 at 21:47

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