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GRE Quadratic Equation
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05 Jan 2016, 21:40
05 Jan 2016, 21:40
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Hi, I cannot seem to figure this question out. I understand quadratic Equations but how do I apply it to this question?
If a ≠ b and a-b , √a - √b , then in terms of b, a equals:
a. √b
b. b
c. b - 2√b + 1
d. b + 2√b + 1
e. b^2 - 2b√b + b
I have read the explanation they provide but cannot seem to figure it out. Any help would be greatly appreciated?
Best Regards
If a ≠ b and a-b , √a - √b , then in terms of b, a equals:
a. √b
b. b
c. b - 2√b + 1
d. b + 2√b + 1
e. b^2 - 2b√b + b
I have read the explanation they provide but cannot seem to figure it out. Any help would be greatly appreciated?
Best Regards
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GRE 1: Q165 V161
Re: GRE Quadratic Equation
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06 Jan 2016, 04:30
06 Jan 2016, 04:30
2
Expert Reply
Hello again,
I will solve the problems step by step:
\(a - b = sqrt a - sqrt b\)
or, \((sqrt a - sqrt b)*(sqrt a + sqrt b) = sqrt a - sqrt b\) since \(a^2 - b^2 = (a-b)(a+b)\)
or, \((sqrt a + sqrt b) = 1\) this part is crucial since its given a not equal to b, so we can cancel the non-zero term \(sqrt a - sqrt b\) on both sides
or, \(sqrt a = 1- sqrt b\) Now squaring both terms
or,\(a =(1- sqrt b)^2\)
or, \(a =1 + b - 2sqrt b\)
So the given answer is choice C.
I will solve the problems step by step:
\(a - b = sqrt a - sqrt b\)
or, \((sqrt a - sqrt b)*(sqrt a + sqrt b) = sqrt a - sqrt b\) since \(a^2 - b^2 = (a-b)(a+b)\)
or, \((sqrt a + sqrt b) = 1\) this part is crucial since its given a not equal to b, so we can cancel the non-zero term \(sqrt a - sqrt b\) on both sides
or, \(sqrt a = 1- sqrt b\) Now squaring both terms
or,\(a =(1- sqrt b)^2\)
or, \(a =1 + b - 2sqrt b\)
So the given answer is choice C.
gmatclubot
Re: GRE Quadratic Equation [#permalink]
06 Jan 2016, 04:30
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