My Math Forum Why does the order of these radicals matter?

 Pre-Calculus Pre-Calculus Math Forum

 May 15th, 2015, 05:12 PM #1 Member   Joined: May 2015 From: U.S.A. Posts: 45 Thanks: 0 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.
 May 15th, 2015, 05:24 PM #2 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 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 $\dfrac{\sqrt{6} - \sqrt{13}}{\sqrt{6} - \sqrt{13}}$
May 15th, 2015, 05:29 PM   #3
Member

Joined: May 2015
From: U.S.A.

Posts: 45
Thanks: 0

Quote:
 Originally Posted by Azzajazz 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

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

$\dfrac{\sqrt{6} + \sqrt{13}}{\sqrt{6} + \sqrt{13}}$ ? 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:

Last edited by Rexan; May 15th, 2015 at 05:59 PM. Reason: Found new information for myself; updating the post

 May 16th, 2015, 01:13 AM #4 Global Moderator   Joined: Dec 2006 Posts: 19,884 Thanks: 1835 $\displaystyle \frac{7}{\sqrt{6} + \sqrt{13}} = \frac{7}{\sqrt{6} + \sqrt{13}}\cdot\frac{\sqrt{6} - \sqrt{13}}{\sqrt{6} - \sqrt{13}} = \frac{7(\sqrt{6} - \sqrt{13})}{-7} = -(\sqrt{6} - \sqrt{13}) = \sqrt{13} - \sqrt{6}$ $\displaystyle \frac{7}{\sqrt{6} + \sqrt{13}} = \frac{7}{\sqrt{6} + \sqrt{13}}\cdot\frac{\sqrt{13} - \sqrt{6}}{\sqrt{13} - \sqrt{6}} = \frac{7(\sqrt{13} - \sqrt{6})}{7} = \sqrt{13} - \sqrt{6}$ Thanks from Rexan
 May 16th, 2015, 04:36 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,502 Thanks: 2511 Math Focus: Mainly analysis and algebra 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. Thanks from Rexan
 May 18th, 2015, 09:47 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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.
May 18th, 2015, 08:45 PM   #7
Senior Member

Joined: Nov 2010
From: Indonesia

Posts: 2,000
Thanks: 132

Math Focus: Trigonometry
Quote:
 Originally Posted by Rexan 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.

May 20th, 2015, 08:50 AM   #8
Member

Joined: May 2015
From: U.S.A.

Posts: 45
Thanks: 0

Quote:
 Originally Posted by Monox D. I-Fly Because if you did, you would summon another irrational number in your denominator.
That clears it up really well. Thanks!

 May 20th, 2015, 10:11 AM #9 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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 $\displaystyle \sqrt{13}- \sqrt{6}$.

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Confused1234 Art 0 February 25th, 2015 10:19 AM shunya Elementary Math 3 July 16th, 2014 06:31 AM ohaider Algebra 6 February 6th, 2012 08:13 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top