
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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July 19th, 2014, 09:17 PM  #1 
Newbie Joined: Jul 2014 From: australia Posts: 16 Thanks: 0  rationalizing surd fractions
Hey y'all! the question should be attached... I tried solving a) with the fraction multiplied by (1+1/√3)/(1+1/√3) to take away the surd in the denominator. is this the right approach? i also tried doing this for b) as well. the answers are: a) 1/2 + 1/2 √3 b) 3 + 2 √2 
July 19th, 2014, 10:10 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,111 Thanks: 1002  
July 19th, 2014, 10:27 PM  #3 
Newbie Joined: Jul 2014 From: australia Posts: 16 Thanks: 0 
i solved it like this: (1/√3)/(11/√3) * (1+1/√3)/(1+1/√3) = [1/√3 (1+1/√3)] / [1(1/√3)^2 = [1/√3 + 1/3] / 1 (1/3) = [1/√3 + 1/3] / (2/3) stuck on the next part. 
July 19th, 2014, 10:47 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,111 Thanks: 1002  
July 19th, 2014, 10:52 PM  #5 
Senior Member Joined: Apr 2014 From: Europa Posts: 575 Thanks: 176 
$\displaystyle \dfrac{\dfrac{1}{\sqrt3}}{1\dfrac{1}{\sqrt3}} \ = \ \dfrac{\dfrac{1}{\sqrt3}}{\dfrac{\sqrt3  1}{\sqrt3}} \ =\ \dfrac{1}{\sqrt3}:\dfrac{\sqrt31}{\sqrt3}\ =\ \dfrac{1}{\cancel{\sqrt3}}\cdot \dfrac{\cancel{\sqrt3}}{\sqrt3  1}\ =\ \dfrac{1}{\sqrt31}\ =\ \dfrac{1}{\sqrt31} \cdot \dfrac{\sqrt3+1}{\sqrt3+1}\ =\ \dfrac{\sqrt3+1}{2}$

July 19th, 2014, 11:09 PM  #6 
Newbie Joined: Jul 2014 From: australia Posts: 16 Thanks: 0  
July 19th, 2014, 11:34 PM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,111 Thanks: 1002  
July 20th, 2014, 12:09 AM  #8 
Newbie Joined: Jul 2014 From: australia Posts: 16 Thanks: 0 
(1/√3)/(11/√3) * (1+1/√3)/(1+1/√3) = [1/√3 (1+1/√3)] / (1  1/√3)(1 + 1/√3) = [1/√3 +1/3)] / (2/3) = [1/√3 +1/3)] /(2/3) * (3/2)/(3/2) = (3/2)[1/√3 +1/3)] / 1 = 3/2√3 + 3/6 = 3/2√3 + 1/2 this is not the answer, as the actual answer is 1/2 + 1/2√3 not sure what i did wrong....? Last edited by jessjans11; July 20th, 2014 at 12:25 AM. 
July 20th, 2014, 12:50 AM  #9  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,111 Thanks: 1002  Quote:
3/2√3 + 1/2 = (6 + 2√3) / (4√3) = (3 + √3) / (2√3) multiply numerator and denominator by √3: = (3√3 + 3) / 6 =3(√3 + 1) / 6 = (√3 + 1) / 2 = √3/2 + 1/2 ... BINGO! Hope your teacher is kinder next time  
July 20th, 2014, 07:24 AM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,111 Thanks: 1002 
So, to perhaps simplify the procedure: let a = √3 Initial expression: (1/a) / (1  1/a) Multiply Numerator and Denominator by (1+ 1/a): (1/a + 1/a^2) / (1  1/a^2) Simplify: [(a + 1) / a^2] / [(a^2  1) / a^2] Simplify: (a + 1) / (a^2  1) Substitute back in: (√3 + 1) / [(√3)^2  1] Simplify: (√3 + 1) / 2 which is same as √3 / 2 + 1/2 

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