
PreCalculus PreCalculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
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:
$\dfrac{\sqrt{6}  \sqrt{13}}{\sqrt{6}  \sqrt{13}}$ 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  
May 16th, 2015, 01:13 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,096 Thanks: 1905 
$\displaystyle \frac{7}{\sqrt{6} + \sqrt{13}} = \frac{7}{\sqrt{6} + \sqrt{13}}\cdot\frac{\sqrt{6}  \sqrt{13}}{\sqrt{6}  \sqrt{13}} = \frac{7(\sqrt{6}  \sqrt{13})}{7} = (\sqrt{6}  \sqrt{13}) = \sqrt{13}  \sqrt{6}$ $\displaystyle \frac{7}{\sqrt{6} + \sqrt{13}} = \frac{7}{\sqrt{6} + \sqrt{13}}\cdot\frac{\sqrt{13}  \sqrt{6}}{\sqrt{13}  \sqrt{6}} = \frac{7(\sqrt{13}  \sqrt{6})}{7} = \sqrt{13}  \sqrt{6}$ 
May 16th, 2015, 04:36 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,551 Thanks: 2556 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.

May 18th, 2015, 09:47 AM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 
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. 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: 2,001 Thanks: 132 Math Focus: Trigonometry  Quote:
 
May 20th, 2015, 08:50 AM  #8 
Member Joined: May 2015 From: U.S.A. Posts: 45 Thanks: 0  
May 20th, 2015, 10:11 AM  #9 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 
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 $\displaystyle \sqrt{13} \sqrt{6}$.


Tags 
matter, order, radicals 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
It's like math and science don't even matter anymore!  Confused1234  Art  0  February 25th, 2015 10:19 AM 
On why it does not matter  shunya  Elementary Math  3  July 16th, 2014 06:31 AM 
Converting radicals to mixed radicals with fractions  ohaider  Algebra  6  February 6th, 2012 08:13 PM 