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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: 29% (02:14) correct 70% (02:10) wrong based on 31 sessions
$$\frac{8x^{21}+12x^{20}-108x^{19}+\sqrt{36x^4}}{2x} = 3x$$

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

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[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
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Re: GRE Math Challenge #3 [#permalink]
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Re: GRE Math Challenge #3 [#permalink]
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Re: GRE Math Challenge #3 [#permalink]
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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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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
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Re: GRE Math Challenge #3-Which of the following values of x [#permalink]
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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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Which of the following values of x satisfy the above equatio [#permalink]
Expert's post
$$\frac{8x^{21}+12x^{20}-108x^{19}+\sqrt{36x^4}}{2x}=3x$$

Which of the following values of x satisfy the above equation?

Indicate all such statements.

A. −6

B. −4.5

C. −3

D. 0

E. 3

F. 4.5

G. 6

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Re: Which of the following values of x satisfy the above equatio [#permalink]
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Solve for x (but attention, x = 0 can't be the solution, because we can't divide by 0)

--> (8x^21 + 12x^20 − 108x^19 + √(36x^4)) / 2x = 3x | *2x
--> 8x^21 + 12x^20 − 108x^19 + 6x^2 = 6x^2 | -6x^2 | :4
--> 2x^21 + 3 x^20 - 27 x^19 = 0 --> but x=0 not possible
--> x^19 (2x^2 + 3x - 27) = 0 --> quadratic formula

--> (-3 +- √(9+216)) / 4 --> x = -4,5 or x = 3 --> B, E
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Re: GRE Math Challenge #3-Which of the following values of x [#permalink]
Expert's post
Bump for further discussion
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Re: Which of the following values of x satisfy the above equatio [#permalink]
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Go ahead dig in and start simplifying!

First, multiply both sides by 2x.

(by the way, that move wouldn't be OK if x were equal to zero, but "x" cannot be equal to 0 because that would render the original equation undefined, but that kind of detail may or may not be necessary to realize in a GRE question like this which demands values)

You'll get 6x^2 on the righthandside, and that happens to be the square root of 36x^4. So, you can subtract 6x^2 from both sides to get this equation:

8x^21 + 12x^20 - 108x^19 = 0

Divide everything by x^19 (remember, you can do this because you know x=/=0)

8x^2 + 12x - 108 = 0

Divide everything by 4

2x^2 + 3x - 27 = 0

Do some trial and error to get this factoring:

(2x + 9) (x - 3) = 0

x = -4.5 and 3. Re: Which of the following values of x satisfy the above equatio   [#permalink] 30 May 2020, 18:01
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