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GRE Math Challenge #3-Which of the following values of x

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GRE Math Challenge #3-Which of the following values of x [#permalink] New post 09 Aug 2014, 02:35
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77% (00:30) correct 22% (03:16) wrong based on 9 sessions
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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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Re: GRE Math Challenge #3 [#permalink] New post 09 Aug 2014, 02:38
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The official answers are B,E.
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Re: GRE Math Challenge #3 [#permalink] New post 04 Sep 2014, 07:13
B,E
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Re: GRE Math Challenge #3 [#permalink] New post 16 May 2015, 13:29
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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

Answers:
[Reveal] Spoiler:
B and E


Cheers,
Brent
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Re: GRE Math Challenge #3-Which of the following values of x [#permalink] New post 21 Aug 2017, 08:57
I quite didn't get how the expression 2x^2+3x-27 =0 was factorised
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Re: GRE Math Challenge #3-Which of the following values of x [#permalink] New post 21 Aug 2017, 09:20
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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.

How about 27 and -1?
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

How about 9 and -3?
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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Re: GRE Math Challenge #3-Which of the following values of x   [#permalink] 21 Aug 2017, 09:20
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GRE Math Challenge #3-Which of the following values of x

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