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If a≠b, and a^2/(a-b)=b, which of the following could be a?

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If a≠b, and a^2/(a-b)=b, which of the following could be a?  [#permalink]

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New post 01 Feb 2018, 01:39
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

68% (02:35) correct 32% (02:43) wrong based on 75 sessions

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[GMAT math practice question]

If \(a≠b\), and \(\frac{a^2}{(a-b)}=b\), which of the following could be \(a\)?

A. -1
B. 0
C. 1
D. 2
E. a does not exist

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If a≠b, and a^2/(a-b)=b, which of the following could be a?  [#permalink]

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New post 01 Feb 2018, 11:37
MathRevolution wrote:
[GMAT math practice question]

If \(a≠b\), and \(\frac{a^2}{(a-b)}=b\), which of the following could be \(a\)?

A. -1
B. 0
C. 1
D. 2
E. a does not exist



i think this question goes well with good wine :)

how should i know what value is b?

start plugging -1 ---> i get 1 =-b+b so how can understand if \(a≠b\) :? cant understand the concept ..
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Re: If a≠b, and a^2/(a-b)=b, which of the following could be a?  [#permalink]

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New post 01 Feb 2018, 23:10
1
MathRevolution wrote:
[GMAT math practice question]

If \(a≠b\), and \(\frac{a^2}{(a-b)}=b\), which of the following could be \(a\)?

A. -1
B. 0
C. 1
D. 2
E. a does not exist


We can write the equation as -

\(b^2 - ab + a^2 = 0\)
\(b = a+- \sqrt{(-3a^2)}.\).. So, there is no value of a for which b will be a non-imaginary number.
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If a≠b, and a^2/(a-b)=b, which of the following could be a?  [#permalink]

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New post 04 Feb 2018, 18:13
=>

\(\frac{a^2}{(a-b)}=b\)
\(⇔ a^2=b(a-b)\)
\(⇔ a^2=ab – b^2\)
\(⇔ a^2 - ab + b^2 = 0\)
Since \(a≠b\), one of a and b is not zero.
If \(a ≠ 0\) or \(b ≠ 0\), \(a^2 - ab + b^2\) cannot be zero for the following reason:
\(a^2 - ab + b^2 = a^2 – 2a(\frac{b}{2}) + b^2 = a2 – 2a(\frac{b}{2}) + \frac{b^2}{3} + (\frac{3}{4})b^2\)
\(= (a – \frac{b}{2})^2 + (\frac{3}{4})b^2 > 0\)
Thus, there is no pair of real numbers \((a,b)\) satisfying the equation \(a^2 - ab + b^2 = 0.\)

Therefore, \(a\) cannot exist, and the answer is E.
Answer: E
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If a≠b, and a^2/(a-b)=b, which of the following could be a? &nbs [#permalink] 04 Feb 2018, 18:13
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