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Director  Joined: 16 May 2014
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GRE Math Challenge #3-Which of the following values of x [#permalink]
Expert's post 00:00

Question Stats: 70% (00:30) correct 30% (02:23) wrong based on 10 sessions
Attachment: q2.png [ 11.35 KiB | Viewed 4285 times ]

Which of the following values of x satisfy the equation?
Indicate all possible values.

A) -6
B) -4.5
C) -3
D) 0
E) 3
F) 4.5
G) 6
[Reveal] Spoiler: OA

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If you find this post helpful, please press the kudos button to let me know !  Director  Joined: 16 May 2014
Posts: 595
GRE 1: Q165 V161 Followers: 97

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Re: GRE Math Challenge #3 [#permalink]
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If you find this post helpful, please press the kudos button to let me know ! Intern Joined: 03 Sep 2014
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Re: GRE Math Challenge #3 [#permalink]
B,E GRE Instructor Joined: 10 Apr 2015
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Re: GRE Math Challenge #3 [#permalink]
2
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Expert's post
soumya1989 wrote:
Attachment:
q2.png

Which of the following values of x satisfy the equation?
Indicate all possible values.

A) -6
B) -4.5
C) -3
D) 0
E) 3
F) 4.5
G) 6

For our first step, let's eliminate the fraction by multiplying both sides by 2x to get:
8x^21 + 12x^20 - 108x^19 + √(36x^4) = 6x^2
Now simplify √(36x^4) to get: 8x^21 + 12x^20 - 108x^19 + 6x^2 = 6x^2
Subtract 6x^2 from both sides to get: 8x^11 + 12x^20 - 108x^19 = 0
Factor out 4x^19 to get: 4x^19(2x^2 + 3x - 27) = 0
Factor to get: 4x^19(2x + 9)(x - 3) = 0
So, it's possible that 4x^19 = 0, in which case x = 0*
Or 2x + 9 = 0, in which case x = -4.5
Or x - 3 = 0, in which case x = 3

IMPORTANT: Early in our approach, we multiplied both sides by 2x. We need to be careful here, because if we inadvertently multiplied both sides by zero, then our solution may be incorrect.
Notice that x = 0 IS NOT a solution to the original equation.
If plug in x = 0 we get: 0/0 = 0, which is not true.
So, x = 0 is NOT a valid solution.

So, our two valid solutions are x = -4.5 and x = 3

[Reveal] Spoiler:
B and E

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com Intern Joined: 20 Aug 2017
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Re: GRE Math Challenge #3-Which of the following values of x [#permalink]
I quite didn't get how the expression 2x^2+3x-27 =0 was factorised GRE Instructor Joined: 10 Apr 2015
Posts: 1968
Followers: 60

Kudos [?]: 1797  , given: 9

Re: GRE Math Challenge #3-Which of the following values of x [#permalink]
2
KUDOS
Expert's post
abbasiime wrote:
I quite didn't get how the expression 2x^2 + 3x - 27 = 0 was factorised

There's a formal technique for factoring quadratics that do NOT have a 1 as the coefficient of the x^2 term.
However, the GRE will never give us a super complicated expression to factor.
So, we can just play around with some numbers.

First, we know that the two x-terms must multiply together to give us 2x^2.
So, one term must be 2x and the other term must be x

So far, we have: 2x^2 + 3x - 27 = (2x +/- ??)(x +/- ??)

For the two missing terms, we know that they multiply to get -27
So, let's try some options.

We get: 2x^2 + 3x - 27 = (2x + 27)(x - 1)
To see if this works, we'll EXPAND (2x + 27)(x - 1) using the FOIL Method
(2x + 27)(x - 1) = 2x^2 - 2x + 27x - 27
= 2x^2 + 25x - 27
This doesn't equal 2x^2 + 3x - 27. So, our missing numbers are NOT 27 and -1

We get: 2x^2 + 3x - 27 = (2x + 9)(x - 3)
To see if this works, we'll EXPAND (2x + 9)(x - 3) using the FOIL Method
(2x + 9)(x - 3) = 2x^2 - 6x + 9x - 27
= 2x^2 + 3x - 27
Perfect - this equals the original expression!
So, our missing numbers are NOT 27 and -1

Go, given: 2x^2 + 3x - 27 = 0
Factor to get: (2x + 9)(x - 3) = 0
This means EITHER 2x + 9 = 0 OR x - 3 = 0

If 2x + 9 = 0, then x = -4.5
If x - 3 = 0, then x = 3

Our solutions are x = -4.5 and x = 3
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Brent Hanneson – Creator of greenlighttestprep.com  Re: GRE Math Challenge #3-Which of the following values of x   [#permalink] 21 Aug 2017, 09:20
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