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Himalayan
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which is equal to ^2 a) 1 b) 5 c) sqrt (6) d) 5 - 04 Jul 2007, 11:55
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which is equal to [1 / {sqrt (3) - sqrt (2)}]^2

a) 1
b) 5
c) sqrt (6)
d) 5 - sqrt (6)
e) 5 + 2 sqrt (6)


https://www.gmatclub.com/phpbb/viewtopi ... 577#339577



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which is equal to ^2 a) 1 b) 5 c) sqrt (6) d) 5 - 04 Jul 2007, 12:23
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[1 / {sqrt (3) - sqrt (2)}]^2

=[3^1/2 + 2^1/2]^2 = 5+2*6^1/2 E.
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which is equal to ^2 a) 1 b) 5 c) sqrt (6) d) 5 - 05 Jul 2007, 02:31
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[1/(3^1/2 - 2^1/2)]^2
= (3^1/2 + 2^1/2)^2
= 3+2+2(3^1/2)(2^1/2)
= 5 + 2(6^1/2)
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which is equal to ^2 a) 1 b) 5 c) sqrt (6) d) 5 - 05 Jul 2007, 02:40
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Are you guys saying that 1/[sqrt(x) - sqrt(y)] = sqrt(x) + sqrt(y) ??!
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ian7777
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which is equal to ^2 a) 1 b) 5 c) sqrt (6) d) 5 - 05 Jul 2007, 05:38
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Mishari
Are you guys saying that 1/[sqrt(x) - sqrt(y)] = sqrt(x) + sqrt(y) ??!
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No, they aren't. But it works out that way IF AND ONLY IF, using the variables you put above, y = x - 1.

If we use numbers, you'll see why.

Let's take 1/(root(6)-root(5))

The way to clear out roots from the denominator of a fraction is to multiply the fraction by one (which doesn't change the value) but using the denominator as the way to it.

As an aside, if we have 2/root(3), we would multiply it by root(3)/root(3), and we'd get 2root(3)/3, and we'd have cleared out the root from the bottom.

In this case we have to be a little more sophisticated. Look what happens when you use the (x+y)(x-y)=x^2 - y^2 rule on this:

Multiply the fraction by (root(6)+root(5))/(root(6)+root(5)).

You will get (root(6)+root(5))/(6-5)

Since 6 - 5 is 1, that's just root(6)+root(5), and so we've made the theory work.

Try it with roots that are not different from one, and it won't work out. But the method is still right even if that's not the case, because the GMAT could have figured some other trick out.

I'll post a similar question.

Hope that wasn't too long winded.
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which is equal to ^2 a) 1 b) 5 c) sqrt (6) d) 5 - 05 Jul 2007, 07:21
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thats wonderful explaination.

guys, just remember this simple step, whenever you see Sqrt(x) - Sqrt(y) in the denominator, multiply the fraction with (sqrt(x) + sqrt(y))/(sqrt(x) + sqrt(y)) and then solve it by applying a^2-b^2 formula. This is very simple method to solve complex roots questions. Hope this helps.



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