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If a^2 – ab = 2b^2, which of the following could be b/a

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If a^2 – ab = 2b^2, which of the following could be b/a [#permalink]

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Could you help me with following question.

If \(a^2\) – \(ab\) = \(2b^2\), which of the following could be b/a ?

I. \(1/2\)
II. \(2\)
III. \(-1\)

(A) I only
(B) II only
(C) I and II
(D) I and III
(E) II and III
[Reveal] Spoiler: OA

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Re: If a^2 – ab = 2b^2, which of the following could be b/a [#permalink]

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New post 31 Jul 2017, 21:59
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\(a^2 – ab = 2b^2\) can be written as \(a^2 – 2b^2 - ab = 0\) --- Equation 1

Divide Equation 1 by \(a^2\), this results in ==> 1 - 2 \((\frac{b}{a})^2\) - \(\frac{b}{a}\) = 0 -- Equation 2

Replace \(\frac{b}{a}\) by x in Equation 2, this results in ==> 1 - 2 \(x^2\) - x = 0 ==> 2 \(x^2\) + x - 1 = 0 ==> (2x-1)(x+1) = 0

Therefore, x = -1 or\(\frac{1}{2}\)

\(\frac{b}{a}\) = -1 or \(\frac{1}{2}\)

Answer is D

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Re: If a^2 – ab = 2b^2, which of the following could be b/a [#permalink]

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New post 31 Jul 2017, 22:00
ammuseeru wrote:
Could you help me with following question.

If \(a^2\) – \(ab\) = \(2b^2\), which of the following could be b/a ?

I. \(1/2\)
II. \(2\)
III. \(-1\)

(A) I only
(B) II only
(C) I and II
(D) I and III
(E) II and III


HI..
If you are seeing an equation in terms of variables and value asked in terms of fraction, try dividing the equation by the variable in denominator..

here divide equation \(a^2\) – \(ab\) = \(2b^2\) by \(a^2..\)..
so \(\frac{a^2-ab}{a^2}=\frac{2b^2}{a^2}.......... 1-\frac{b}{a}=2*\frac{b}{a}^2\)..
here you can do two things ..
(A) substitute the given values ..
\(1-\frac{b}{a}=2*\frac{b}{a}^2\)..
1) 1/2..
\(1-\frac{1}{2}=2*\frac{1}{2}^2........1-1/2=1/2\)...YES
2) 2..
\(1-2=2*2^2\).....No
3) -1
\(1-(-1)=2*(-1)^2....1+1=2\).....YES

so I and III
D

(B) solve ..
\(1-\frac{b}{a}=2*\frac{b}{a}^2\)..
let b/a = x..
so \(1-x=2x^2.....2x^2+x-1=0....(2x-1)(x+1)=0\)..
so x=1/2 or x=-1

D
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: If a^2 – ab = 2b^2, which of the following could be b/a [#permalink]

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New post 01 Aug 2017, 09:51
quantumliner wrote:
\(a^2 – ab = 2b^2\) can be written as \(a^2 – 2b^2 - ab = 0\) --- Equation 1

Divide Equation 1 by \(a^2\), this results in ==> 1 - 2 \((\frac{b}{a})^2\) - \(\frac{b}{a}\) = 0 -- Equation 2

Replace \(\frac{b}{a}\) by x in Equation 2, this results in ==> 1 - 2 \(x^2\) - x = 0 ==> 2 \(x^2\) + x - 1 = 0 ==> (2x-1)(x+1) = 0

Therefore, x = -1 or\(\frac{1}{2}\)

\(\frac{b}{a}\) = -1 or \(\frac{1}{2}\)

Answer is D


How did you factorize into 2 \(x^2\) + x - 1 = 0 ==> (2x-1)(x+1) = 0

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Re: If a^2 – ab = 2b^2, which of the following could be b/a [#permalink]

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New post 02 Aug 2017, 10:24
ammuseeru wrote:
quantumliner wrote:
\(a^2 – ab = 2b^2\) can be written as \(a^2 – 2b^2 - ab = 0\) --- Equation 1

Divide Equation 1 by \(a^2\), this results in ==> 1 - 2 \((\frac{b}{a})^2\) - \(\frac{b}{a}\) = 0 -- Equation 2

Replace \(\frac{b}{a}\) by x in Equation 2, this results in ==> 1 - 2 \(x^2\) - x = 0 ==> 2 \(x^2\) + x - 1 = 0 ==> (2x-1)(x+1) = 0

Therefore, x = -1 or\(\frac{1}{2}\)

\(\frac{b}{a}\) = -1 or \(\frac{1}{2}\)

Answer is D


How did you factorize into 2 \(x^2\) + x - 1 = 0 ==> (2x-1)(x+1) = 0


2\(x^2\) + x - 1 = 0 can be written as 2\(x^2\) - x + 2x - 1 = 0 ==> x(2x-1) +1(2x-1) ==> (2x-1)(x+1)

Hope this helps

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Re: If a^2 – ab = 2b^2, which of the following could be b/a   [#permalink] 02 Aug 2017, 10:24
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