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July 19th, 2014, 09:17 PM   #1
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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.

a) 1/2 + 1/2 √3
b) 3 + 2 √2
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July 19th, 2014, 10:10 PM   #2
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Quote:
 Originally Posted by jessjans11 is this the right approach?
YES.

(√3 - 1) * (√3 + 1) = 2 : you knew that?

 July 19th, 2014, 10:27 PM #3 Newbie   Joined: Jul 2014 From: australia Posts: 16 Thanks: 0 i solved it like this: (1/√3)/(1-1/√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
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Quote:
 Originally Posted by jessjans11 (1/√3)/(1-1/√3) * (1+1/√3)/(1+1/√3) = [1/√3 (1+1/√3)] / [1-(1/√3)^2
Your denominator [1-(1/√3)^2 is not correct; should be:
[(1 - 1/√3)(1 + 1/√3)]

 July 19th, 2014, 10:52 PM #5 Senior Member     Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177 $\displaystyle \dfrac{\dfrac{1}{\sqrt3}}{1-\dfrac{1}{\sqrt3}} \ = \ \dfrac{\dfrac{1}{\sqrt3}}{\dfrac{\sqrt3 - 1}{\sqrt3}} \ =\ \dfrac{1}{\sqrt3}:\dfrac{\sqrt3-1}{\sqrt3}\ =\ \dfrac{1}{\cancel{\sqrt3}}\cdot \dfrac{\cancel{\sqrt3}}{\sqrt3 - 1}\ =\ \dfrac{1}{\sqrt3-1}\ =\ \dfrac{1}{\sqrt3-1} \cdot \dfrac{\sqrt3+1}{\sqrt3+1}\ =\ \dfrac{\sqrt3+1}{2}$ Thanks from jessjans11
July 19th, 2014, 11:09 PM   #6
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Quote:
 Originally Posted by Denis Your denominator [1-(1/√3)^2 is not correct; should be: [(1 - 1/√3)(1 + 1/√3)]
if you solve (1 - 1/√3)(1 + 1/√3) wouldnt it still equal 2/3

July 19th, 2014, 11:34 PM   #7
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Quote:
 Originally Posted by jessjans11 if you solve (1 - 1/√3)(1 + 1/√3) wouldnt it still equal 2/3
Yes.
But [1-(1/√3)]^2 that you (seem to) show is not.

 July 20th, 2014, 12:09 AM #8 Newbie   Joined: Jul 2014 From: australia Posts: 16 Thanks: 0 (1/√3)/(1-1/√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
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Quote:
 Originally Posted by jessjans11 = 3/2√3 + 1/2 this is not the answer, as the actual answer is 1/2 + 1/2√3
You're doing fine, Jess!

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: 14,415 Thanks: 1025 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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