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Wierd sqrt problem...

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by just4myself » Mon Aug 25, 2008 9:01 am
Which of the following is equal to [1/ (sqrt (3)-sqrt(2))]^2
A) 1
B) 5
C) SQRT (6)
D) 5-SQRT (6)
E) 5+2SQRT(6)

Please elaborate...

Thank you...
Last edited by just4myself on Mon Aug 25, 2008 9:14 am, edited 1 time in total.
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by pepeprepa » Mon Aug 25, 2008 9:09 am
Use approximations of the square root.
You should know at least sqrt2 and sqrt3
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by just4myself » Mon Aug 25, 2008 9:17 am
Yeah i tried ur method...
sqrt(2) ~ 1.4
sqrt(3) ~ 1.7

The eqn is [1/ (sqrt (3)-sqrt(2))]^2
So, [1/ (1.7-1.4)]^2
.... [1/0.3^2]
.... [1/0.09]...
What after this?

Btw the OA is E...
Any explanations
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by pepeprepa » Mon Aug 25, 2008 9:24 am
1/0.09=1/(9/100)=100/9 (let's say little more than 10)

A) 1
Out
B) 5
Out
D) 5-SQRT (6)
Out

C) SQRT (6)
sqrt(6)=2.45
E) 5+2SQRT(6)
So that's the only one
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Re: Wierd sqrt problem...

by Ian Stewart » Mon Aug 25, 2008 9:58 am
just4myself wrote:Which of the following is equal to [1/ (sqrt (3)-sqrt(2))]^2
A) 1
B) 5
C) SQRT (6)
D) 5-SQRT (6)
E) 5+2SQRT(6)

Please elaborate...

Thank you...
With roots in the denominator, you will want to rationalize, using the difference of squares. Notice that:

1/ (sqrt (3)-sqrt(2))
= [1/ (sqrt (3)-sqrt(2))]*[(sqrt(3)+sqrt(2))/(sqrt(3)+sqrt(2))]
= sqrt(3)+sqrt(2)

The question asks for the square of this:

(sqrt(3)+sqrt(2))^2 = 3 + 2*sqrt(6) + 2 = 5 + 2sqrt(6)

I don't see anything wrong with using estimates for sqrt(2) and sqrt(3) for this particular question, but I do think it's better to understand the method above- you might instead see a question with much larger roots that you are not able to estimate quickly.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by just4myself » Mon Aug 25, 2008 10:20 am
But Ian,
In the given question,
(sqrt(3)+sqrt(2))^2 = 3 + 2*sqrt(6) + 2 = 5 + 2sqrt(6)
is the denominator right?
So we cannot get the answer directly...
Now in fact the ans E becomes more ambiguous
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by Ian Stewart » Mon Aug 25, 2008 11:43 am
just4myself wrote:But Ian,
In the given question,
(sqrt(3)+sqrt(2))^2 = 3 + 2*sqrt(6) + 2 = 5 + 2sqrt(6)
is the denominator right?
So we cannot get the answer directly...
Now in fact the ans E becomes more ambiguous
I'm not sure what you mean by 'we cannot get the answer directly'; we can, as I outlined above. The denominator is indeed right- the denominator is equal to 1. We multiplied the top and bottom of the fraction by sqrt(3) + sqrt(2). In the denominator, we therefore have:

(sqrt(3) + sqrt(2))(sqrt(3) - sqrt(2))

Notice this is just the factorization of a difference of squares. When you multiply this out, you get:

(sqrt(3))^2 - (sqrt(2))^2 = 3 - 2 = 1
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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