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It was really a long time since I did math at school and now I'm preparing for a admission test to college. I took a look at the older tests and immediately got stuck at the first question. It really feels like I have to shape up. Anyway so here's the equation:

$$x = \frac{\frac{a-b}{\sqrt a+\sqrt b}+\sqrt b}{\frac{a-b}{\sqrt a+\sqrt b}-\sqrt a}$$

...and it is supposed to end up in $-\sqrt\frac{a}{b}$

I tried a few different ways of doing it, but always end up wrong. I don't wanna post my attempts here since I honestly don't seem to know what I'm doing. Also this is my first post here on math.stackexchange.com so please tell me if I'm doing something wrong. Any help is greatly appreciated (with the problem).

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    $\begingroup$ I don't wanna post my attempts here since I honestly don't seem to know what I'm doing WRONG! You have to post them, because that's the only way we can help you. $\endgroup$ – 5xum Nov 7 '16 at 13:37
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    $\begingroup$ If you multiply by $\sqrt a+\sqrt b$ you get $\frac {a-b+\sqrt {ab}+b}{a-b-a-\sqrt {ab}}$ Now simplify and clear the radical from the denominator. $\endgroup$ – lulu Nov 7 '16 at 13:43
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    $\begingroup$ Nested fractions are annoying. Get rid of them first. $\endgroup$ – Sophie Nov 7 '16 at 13:43
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Notice that $$a-b = (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})$$

so you get $$(\sqrt{a}-\sqrt{b}+\sqrt{b})/(\sqrt{a}-\sqrt{b}-\sqrt{a})=-\sqrt{a\over b}$$

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Hint:

Multiply the denominator and numerator by $\sqrt{a} + \sqrt b$. Then, tell us what you get, and we can talk further.

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Let us start by simplifying the numerator and denominator separately:

$$num = \frac{a-b}{\sqrt a+\sqrt b}+\sqrt{b} = \frac{a-b}{\sqrt a+\sqrt b} + \frac{\sqrt{b}({\sqrt a+\sqrt b})}{{\sqrt a+\sqrt b}}$$ by dividing and multiplying $\sqrt{b}$ by the denominator of the first term. Now we expand and do the maths:

$$num = \frac{a-b+\sqrt{b}{(\sqrt{a}+\sqrt{b})}}{\sqrt{a}+\sqrt{b}} = \frac{a-b+\sqrt{ab}{+b}}{\sqrt{a}+\sqrt{b}} = \frac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}$$

By a similar process we get that the denominator is

$$den = \frac{a-b}{\sqrt a+\sqrt b}-\sqrt{a} = \frac{-b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}$$

Now when you put them back together, you use the following rule: $$\frac{\frac{a}{d}}{\frac{b}{d}} = \frac{a}{b}$$ to get

$$x = \frac{a+\sqrt{ab}}{-b-\sqrt{ab}}$$

Now try spliting $a$ into $\sqrt{a}\sqrt{a}$ and $-b$ into $-\sqrt{b}\sqrt{b}$ and use some algebraic manipulation to get to the final result.

EDIT: as per request, I will go further. (Notice that I had a very unfortunate sign flipped on the simplification of the denominator above)

$a = \sqrt{a}\sqrt{a}$ and $-b = -\sqrt{b}\sqrt{b}$ so that, substituting back,

$$x = \frac{\sqrt{a}\sqrt{a}+\sqrt{ab}}{-\sqrt{b}\sqrt{b}-\sqrt{ab}} = \frac{\sqrt{a}(\sqrt{a} + \sqrt{b})}{-\sqrt{b}(\sqrt{b} + \sqrt{a})} = \frac{\sqrt{a}}{-\sqrt{b}} = -\sqrt{\frac{a}{b}}$$

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  • $\begingroup$ Hi....this was a while ago but could you please show me how this "algebraic manipulation" works? $\endgroup$ – splokk23 Aug 25 at 0:08
  • $\begingroup$ @splokk23 I edited my answer. I am also very sorry but I had a sign flipped in the first part of my answer. If this answer of mine, or any other answer posted here, solves your problem, please mark it as accepted! Thanks $\endgroup$ – RGS Aug 25 at 8:35
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just simplify the equation:

$ \Large x =\frac{\frac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(a-b\right)}+\sqrt{b}}{\frac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(a-b\right)}-\sqrt{a}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)-\sqrt{a}}=\frac{\sqrt{a}}{-\sqrt{b}}=-\sqrt{\frac{a}{b}}$

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Doing the common denominator and simplifying, I obtain: $\frac{a+\sqrt{ab}}{-(b+\sqrt{ab})}$. From this, I multiply the numerator and the denominator by $b-\sqrt{ab}$ and I get: $\frac{(a+\sqrt{ab})(b-\sqrt{ab})}{-(b+\sqrt{ab})(b-\sqrt{ab})}=\frac{ab-a\sqrt{ab}+b\sqrt{ab}-ab}{-(b^2-b\sqrt{ab}+b\sqrt{ab}-ab)}=\frac{\sqrt{ab}(b-a)}{-b(b-a)}=-\frac{\sqrt{ab}}{b}=-\sqrt{\frac{ab}{b^2}}=-\sqrt{\frac{a}{b}}$

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