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

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Manager
Joined: 17 Mar 2014
Posts: 152

Kudos [?]: 77 [1], given: 353

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

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31 Jul 2017, 13:09
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Question Stats:

64% (01:32) correct 36% (01:26) wrong based on 66 sessions

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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

Kudos [?]: 77 [1], given: 353

Senior Manager
Joined: 24 Apr 2016
Posts: 335

Kudos [?]: 188 [4], given: 48

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

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31 Jul 2017, 21:59
4
KUDOS
$$a^2 – ab = 2b^2$$ 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 $$\frac{1}{2}$$

Kudos [?]: 188 [4], given: 48

Math Expert
Joined: 02 Aug 2009
Posts: 5212

Kudos [?]: 5851 [0], given: 117

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

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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

Kudos [?]: 5851 [0], given: 117

Manager
Joined: 17 Mar 2014
Posts: 152

Kudos [?]: 77 [0], given: 353

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

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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 $$\frac{1}{2}$$

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

Kudos [?]: 77 [0], given: 353

Senior Manager
Joined: 24 Apr 2016
Posts: 335

Kudos [?]: 188 [0], given: 48

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

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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 $$\frac{1}{2}$$

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

Kudos [?]: 188 [0], given: 48

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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