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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.

the answers are:
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 View Post
is this the right approach?
YES.

(√3 - 1) * (√3 + 1) = 2 : you knew that?
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July 19th, 2014, 10:27 PM   #3
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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.
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July 19th, 2014, 10:47 PM   #4
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Quote:
Originally Posted by jessjans11 View Post
(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)]
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July 19th, 2014, 10:52 PM   #5
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$\displaystyle \dfrac{\dfrac{1}{\sqrt3}}{1-\dfrac{1}{\sqrt3}} \ = \ \dfrac{\dfrac{1}{\sqrt3}}{\dfrac{\sqrt3 - 1}{\sqrt3}} \ =\ \dfrac{1}{\sqrt3}:\dfrac{\sqrt3-1}{\sqrt3}\ =\ \dfrac{1}{\cancel{\sqrt3}}\cdot \dfrac{\cancel{\sqrt3}}{\sqrt3 - 1}\ =\ \dfrac{1}{\sqrt3-1}\ =\ \dfrac{1}{\sqrt3-1} \cdot \dfrac{\sqrt3+1}{\sqrt3+1}\ =\ \dfrac{\sqrt3+1}{2}$
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July 19th, 2014, 11:09 PM   #6
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Quote:
Originally Posted by Denis View Post
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
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July 19th, 2014, 11:34 PM   #7
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Quote:
Originally Posted by jessjans11 View Post
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.
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July 20th, 2014, 12:09 AM   #8
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(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.
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July 20th, 2014, 12:50 AM   #9
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Quote:
Originally Posted by jessjans11 View Post
= 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
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July 20th, 2014, 07:24 AM   #10
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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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