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?
if a>b>0, then sqrt a^2 + b^2
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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!
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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I am facing same problem.. Even I choose option C...
I am not getting whats wrong with C..
Any help would be appreciable.
I am not getting whats wrong with C..
Any help would be appreciable.
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)
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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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.
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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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
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