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May 15th, 2015, 05:12 PM   #1
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Question Why does the order of these radicals matter?

I'm reviewing slews of math for an upcoming college placement exam, and there's a practice question that demands I rationalize the denominator. The statement is thus:

Rationalize the denominator.

7 divided by (square root of 6 plus square root of 13)

$\displaystyle \frac {7}{\sqrt{6} + \sqrt{13}}$

I guessed it was only asking about the denominator and that the numerator didn't matter (at least that's what I guessed after looking at the options, since there aren't any fractions there), so I figured, out of the options I was given,
√13 + √6
would suffice.

Turns out the answer is √13 - √6 . How does this make sense??

Last edited by Rexan; May 15th, 2015 at 05:31 PM.
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May 15th, 2015, 05:24 PM   #2
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Do you know what rationalising the denominator means? You should have manipulated the fraction until there were no radicals in the denominator.

Start by multiplying the fraction by $\dfrac{\sqrt{6} - \sqrt{13}}{\sqrt{6} - \sqrt{13}}$
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May 15th, 2015, 05:29 PM   #3
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Quote:
Originally Posted by Azzajazz View Post
Do you know what rationalising the denominator means? You should have manipulated the fraction until there were no radicals in the denominator.

Start by multiplying the fraction by $\dfrac{\sqrt{6} - \sqrt{13}}{\sqrt{6} - \sqrt{13}}$
Right. I looked up rationalizing and reviewed that (at least at a very low level I guess), but I'm still getting confused. Why multiply it by

$\dfrac{\sqrt{6} - \sqrt{13}}{\sqrt{6} - \sqrt{13}}$

instead of

$\dfrac{\sqrt{6} + \sqrt{13}}{\sqrt{6} + \sqrt{13}}$ ? Sorry. I haven't been in math class for a while...

Edit:
Okay, I did the multiplying and got the answer, but after redoing the problem and making the order so that the $\displaystyle \sqrt{13}$ came before the $\displaystyle \sqrt{6}$. In the future, would I just have to know to put the larger number before the smaller?

Another edit:
'Kay, conjugates. Totally forgot about those. [link to YouTube]

Last edited by Rexan; May 15th, 2015 at 05:59 PM. Reason: Found new information for myself; updating the post
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May 16th, 2015, 01:13 AM   #4
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$\displaystyle \frac{7}{\sqrt{6} + \sqrt{13}} = \frac{7}{\sqrt{6} + \sqrt{13}}\cdot\frac{\sqrt{6} - \sqrt{13}}{\sqrt{6} - \sqrt{13}} = \frac{7(\sqrt{6} - \sqrt{13})}{-7} = -(\sqrt{6} - \sqrt{13}) = \sqrt{13} - \sqrt{6}$

$\displaystyle \frac{7}{\sqrt{6} + \sqrt{13}} = \frac{7}{\sqrt{6} + \sqrt{13}}\cdot\frac{\sqrt{13} - \sqrt{6}}{\sqrt{13} - \sqrt{6}} = \frac{7(\sqrt{13} - \sqrt{6})}{7} = \sqrt{13} - \sqrt{6}$
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May 16th, 2015, 04:36 AM   #5
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The order only matters in terms of getting the value of the expression correct. $-\sqrt6 + \sqrt{13}$ works too, although it is a slightly odd way to write it.
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May 18th, 2015, 09:47 AM   #6
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The reason why this works is the fact that $\displaystyle (a+ b)(a- b)= a^2- b^2$.
To "rationalize" $\displaystyle \sqrt{6}$, you need to square it: $\displaystyle (\sqrt{6})^2= 6$. And, obviously, the same thing works for $\displaystyle \sqrt{13}$: $\displaystyle (\sqrt{13})^2= 13$.

So to "rationalize" $\displaystyle \sqrt{6}+ \sqrt{13}$, you multiply by $\displaystyle \sqrt{6}- \sqrt{13}$: $\displaystyle (\sqrt{6}+ \sqrt{13})(\sqrt{6}- \sqrt{13})= ((\sqrt{6})^2- (\sqrt{13})^2)= 6- 13= -7$.

Thus, $\displaystyle \sqrt{6}- \sqrt{13}$ answers the question "what must you multiply the numerator and denominator of this fraction by in order to rationalize the fraction?"

IF the problem was really "rationalize the denominator of $\displaystyle \frac{7}{\sqrt{6}+ \sqrt{13}}$", then the solution skipjack gave is correct.
Thanks from topsquark and Rexan

Last edited by skipjack; May 19th, 2015 at 12:14 AM.
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May 18th, 2015, 08:45 PM   #7
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Quote:
Originally Posted by Rexan View Post
Right. I looked up rationalizing and reviewed that (at least at a very low level I guess), but I'm still getting confused. Why multiply it by

$\dfrac{\sqrt{6} - \sqrt{13}}{\sqrt{6} - \sqrt{13}}$

instead of

$\dfrac{\sqrt{6} + \sqrt{13}}{\sqrt{6} + \sqrt{13}}$ ? Sorry. I haven't been in math class for a while...
Because if you did, you would summon another irrational number in your denominator.
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May 20th, 2015, 08:50 AM   #8
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Quote:
Originally Posted by Monox D. I-Fly View Post
Because if you did, you would summon another irrational number in your denominator.
That clears it up really well. Thanks!
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May 20th, 2015, 10:11 AM   #9
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Notice that this has nothing to do with the "order of the radicals". It is that "-" that is important. You could use $\displaystyle \sqrt{6}- \sqrt{13}$ or $\displaystyle \sqrt{13}- \sqrt{6}$.
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