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Sqrt(17+sqrt264) = sqrt(a) + sqrt(b)

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by Brent@GMATPrepNow » Sun Jan 25, 2009 3:06 pm
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Please note that this is not an official GMAT question; it’s my attempt to create difficult (650+ level) GMAT-style questions for this forum.
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Re: Sqrt(17+sqrt264) = sqrt(a) + sqrt(b)

by piyush_nitt » Sun Jan 25, 2009 4:17 pm
Brent Hanneson wrote:Image

Please note that this is not an official GMAT question; it’s my attempt to create difficult (650+ level) GMAT-style questions for this forum.
Sqrt(17+sqrt264) = sqrt(a) + sqrt(b)

squaring both sides

17+sqrt(264) = a + b + 2*sqrt(a)*(sqrt(b)

Comparing ll'ar parts

a + b = 17 - (1)
sqrt(264) = 2*sqrt(a)*sqrt(b) (2)

squaring 2 again

264 = 4*a*b

ab = 66 - (3)

solving 1 and 3

a =11 , b = 6

hence a-b = 5 , IMO E
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by Brent@GMATPrepNow » Sun Jan 25, 2009 4:26 pm
Nice solution piyush. The answer is E.

Here's my solution as well.


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by awesomeusername » Sun Jan 25, 2009 9:29 pm
For some reason, I'm not understanding why we can say 17 = a+b and sqrt(264) = 2sqrt(ab).
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by Brent@GMATPrepNow » Sun Jan 25, 2009 10:44 pm
awesomeusername wrote:For some reason, I'm not understanding why we can say 17 = a+b and sqrt(264) = 2sqrt(ab).
After squaring both sides of the equation we have: 17 + sqrt(264) = a+b+2sqrt(ab)

On both sides, we have terms that are inside the square root and terms outside the square root.

We can we can assume that the terms inside the square root are equal: sqrt(264)=2sqrt(ab) or sqrt(264)=sqrt(4ab)

And we can we can assume that the terms outside the square root are equal: a+b = 17
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by awesomeusername » Mon Jan 26, 2009 6:38 am
Thanks Brent, I think I understand it now
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