# Check my answers to these problems on rationalizing denominators? [closed]

I just need to check my answers for these problems. I did the work already on paper. I'm not trying to get free answers because I know that would hurt me in the long run. So here's the questions:

Rationalize the denominator:

1. $\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}$

2. $\dfrac{(\sqrt{2}-3)^2}{(\sqrt{2}+3)^2}$

3. $\dfrac{\sqrt{a+b}-\sqrt{a-b}}{\sqrt{a+b}+\sqrt{a-b}}$

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## closed as too localized by Michael Greinecker♦, Micah, Joe, Sasha, PaulApr 19 '13 at 4:41

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If you want to check your answers, you ought to show what you found first. This site isn't supposed to simply answer homework questions, so you'll have to show something of your own efforts when you ask for help. –  RecklessReckoner Apr 18 '13 at 21:58
Is that what you intended? By the way, you can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. –  Zev Chonoles Apr 18 '13 at 22:01
Well, the OP apparently wants to check whether (s)he wrote the questions correctly. Yes, you did...I guess. –  DonAntonio Apr 18 '13 at 22:24

Strategy for approaching these kinds of problems:

Let's use $(1)$ as an example:

1. $\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}$

Try multiplying both the numerator and the denominator by the conjugate of the denominator: $$(\sqrt x- \sqrt y) \quad \overset{\large \text conjugate} \iff \quad (\sqrt x + \sqrt y)$$

So, following this strategy, we multiply the numerator and denominator by $\sqrt a - \sqrt b$, which is the conjugate of $\sqrt a + \sqrt b$.

Then we just need to simplify: $$\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}} = \dfrac{(\sqrt a - \sqrt b)\cdot(\sqrt a - \sqrt b)}{(\sqrt a + \sqrt b)\cdot (\sqrt a - \sqrt b)} = \quad ?$$

You'll find that you'll obtain a difference of squares in the denominator, with no radicals!:

$$\dfrac{(\sqrt a - \sqrt b)\cdot(\sqrt a - \sqrt b)}{(\sqrt a + \sqrt b)\cdot (\sqrt a - \sqrt b)} = \dfrac{(\sqrt a)^2 - 2\sqrt{ab} + (\sqrt b)^2}{(\sqrt a)^2 - (\sqrt b)^2} = \dfrac {a - 2\sqrt{ab} + b}{a - b}$$

See if you can generalize this technique and apply it to problems $(2)$ and $(3)$.

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Thanks, dear friend! ;-) Good to see you! I will sleep much better now, having "seen" you ;-) –  amWhy Apr 20 '13 at 3:37
Be peace upon you Amy in the Heaven. Try to come back very soon, we are waiting for you here ;-) –  Babak S. Apr 20 '13 at 3:40
Not quite ready to sleep, but soon I will drift aloft in my slumber ;D –  amWhy Apr 20 '13 at 3:41
1. $\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}} = \dfrac{(a - 2\sqrt{ab} + b)}{a - b}$

2. $\dfrac{(\sqrt{2}-3)^2}{(\sqrt{2}+3)^2} = \dfrac{(\sqrt{2}-3)^4}{49}$

3. $\dfrac{\sqrt{a+b}-\sqrt{a-b}}{\sqrt{a+b}+\sqrt{a-b}} = \dfrac{(\sqrt{a+b}-\sqrt{a-b})^2}{2b}$

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