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if a>b>0, then sqrt a^2 + b^2

Expert replies
by dlamanti » Wed Aug 05, 2009 3:08 pm
Most of the questions from the official GMAT download are posted, but I could find this one. Any help explaining how to do this would be great. My incorrect answer is selected; the correct answer has the box around it.

Is it just factored as a quadratic equation? Do the inequalities give you any info needed for solving the problem?
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Source: — Problem Solving |

by adssaini » Wed Aug 05, 2009 6:26 pm
use formula of a^2 - b^2

its (a+b)(a-b)
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by luiscarlos59 » Tue Mar 22, 2011 8:19 pm
plug in!!

a=6
b=3

on sqrt a^2-b^2
sqrt 36-9
sqrt 27
3 sqrt 3

on statement D
(sqrt 9) * (sqrt 3)
3 sqrt 3

same answer!
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by swapneilghodke » Sat Nov 17, 2012 9:08 am
I am facing same problem.. Even I choose option C...
I am not getting whats wrong with C..
Any help would be appreciable.
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by jkaustubh » Sat Nov 17, 2012 10:30 pm
@ghodke sahab

sqrt[(a-b)^2-2ab]

=sqrt[a^2+b^2-4ab]

which is nowhere equal to a^2-b^2
Replying a query takes patience and time. The least a person can do is to thank the reply.
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by unknown13 » Sun Nov 18, 2012 1:56 am
Hi
I think its simple; this uses the below mention formula
(a^2-b^2)
=(a-b)*(a+b)

thanks and regards
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by talueng » Sun Nov 18, 2012 2:25 am
Just want to emphasize that you should memorize the basic factoring rules such as (helpful in this example):
a²-b² = (a+b)(a-b)

Remembering (and being able to apply) these rules is not only handy (as it helps to reduce time spend and errors); I would even say they are essential for some questions. And, after all, there aren't that many to remember:

a²-b² = (a+b)(a-b)
(a+b)² = a²+2ab+b²
(a-b)² = a²-2ab+b²
a²+(p+q)a+pq = (a+p)(a+q)
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by StrawberryCow » Fri Apr 11, 2014 5:55 am
a>b>0
Let a=2 and b=1

So, sqrt a^2-b^2 = sqrt 4-1 = sqrt 3

Now run through the answer choices
A. a+b - sqrt 2ab --> 3 - sqrt 2x2 = 3 - 2 = 1 -- Does Not Match sqrt 3
B. a - b + sqrt 2ab --> 1-2 = -1 -- Does Not Match sqrt 3
C. sqrt (a - b)^2 - 2ab --> sqrt -3 -- Does Not Match sqrt 3
D. (sqrt a + b) (sqrt a - b) --> sqrt 3 x sqrt 1
now , sqrt 1 = 1 , so matches sqrt 3 as solved initially.
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by Brent@GMATPrepNow » Fri Apr 11, 2014 8:07 am
There's a nice rule that says √(xy) = (√x)(√y)
For example √(16*49) = (√16)(√49) = (4)(7) = 28

Since a² - b² = (a + b)(a - b), can use the above rule to conclude that...
its √[a² - b²] = √[(a+b)(a-b)] = [√(a+b)][√(a-b)] = D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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