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GRE Math Challenge #3-Which of the following values of x
[#permalink]
09 Aug 2014, 02:35

2

Expert Reply

9

Bookmarks

Question Stats:

\(\frac{8x^{21}+12x^{20}-108x^{19}+\sqrt{36x^4}}{2x} = 3x\)

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

Kudos for the right answer and explanation

_________________

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

Kudos for the right answer and explanation

_________________

My GRE Resources

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If you find this post helpful, please press the kudos button to let me know !

Free GRE resources | GRE Prep Club Quant Tests

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Re: GRE Math Challenge #3
[#permalink]
16 May 2015, 13:29

4

Expert Reply

1

Bookmarks

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:

Show: ::

B and E

Cheers,

Brent

_________________

Re: GRE Math Challenge #3
[#permalink]
09 Aug 2014, 02:38

1

Expert Reply

The official answers are B,E.

_________________

My GRE Resources

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If you find this post helpful, please press the kudos button to let me know !

_________________

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Re: GRE Math Challenge #3-Which of the following values of x
[#permalink]
21 Aug 2017, 08:57

I quite didn't get how the expression 2x^2+3x-27 =0 was factorised

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

2

Expert Reply

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

_________________

Which of the following values of x satisfy the above equatio
[#permalink]
01 Feb 2020, 11:51

Expert Reply

\(\frac{8x^{21}+12x^{20}-108x^{19}+\sqrt{36x^4}}{2x}=3x\)

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

Kudos for the right answer and explanation

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

Kudos for the right answer and explanation

Re: Which of the following values of x satisfy the above equatio
[#permalink]
03 Feb 2020, 08:43

1

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

--> (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

Re: GRE Math Challenge #3-Which of the following values of x
[#permalink]
21 Feb 2020, 05:12

Expert Reply

Bump for further discussion

Re: Which of the following values of x satisfy the above equatio
[#permalink]
30 May 2020, 18:01

2

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.

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]
27 Jul 2020, 02:10

It took a lot of time

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Re: Which of the following values of x satisfy the above equatio
[#permalink]
15 Nov 2021, 12:21

GreenlightTestPrep wrote:

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:

Show: ::

B and E

Cheers,

Brent

Hi, How you multiply by 2x? since we don't know whether x=0

Which of the following values of x satisfy the above equatio
[#permalink]
16 Nov 2021, 00:58

1

\(\frac{8x^{21}+12x^{20}-108x^{19}+\sqrt{36x^4}}{2x} = 3x\)

\(8x^{21} + 12x^{20} - 108x^{19} + 6x^2 = 3x(2x) = 6x^2\)

\(8x^{21} + 12x^{20} - 108x^{19} = 0\)

\(x^{19}(8x^2 + 12x - 108) = 0\)

\(x^{19} = 0\) OR \(8x^2 + 12x - 108 = 0\)

\(8x^2 + 12x - 108 = 0\)

\(4(2x^2 + 3x - 27) = 0\)

\(2x^2 + 3x - 27 = 0\)

Using formula \(x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}\)

\(x = \frac{-3 ± \sqrt{3^2 - 4(2)(-27)}}{2(2)} = \frac{-3 ± \sqrt{225}}{2(2)} = \frac{-3 ± 15}{4} = \frac{12}{4} and \frac{-18}{4} = \textbf{3} and \textbf{-4.5}\)

Now we also get one more solution of \(x = 0\), but if you put that value in the equation, we get LHS as \(\frac{0}{0}\), which in language of Mathematics is defined as Undetermined, meaning its a value that cannot be determined based on the current rules of Mathematics. Another such Undetermined is \(\frac{\infty}{\infty}\)

Hence, \(x = 0\) cannot be a possible solution

Hence, Answer are B and E

\(8x^{21} + 12x^{20} - 108x^{19} + 6x^2 = 3x(2x) = 6x^2\)

\(8x^{21} + 12x^{20} - 108x^{19} = 0\)

\(x^{19}(8x^2 + 12x - 108) = 0\)

\(x^{19} = 0\) OR \(8x^2 + 12x - 108 = 0\)

\(8x^2 + 12x - 108 = 0\)

\(4(2x^2 + 3x - 27) = 0\)

\(2x^2 + 3x - 27 = 0\)

Using formula \(x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}\)

\(x = \frac{-3 ± \sqrt{3^2 - 4(2)(-27)}}{2(2)} = \frac{-3 ± \sqrt{225}}{2(2)} = \frac{-3 ± 15}{4} = \frac{12}{4} and \frac{-18}{4} = \textbf{3} and \textbf{-4.5}\)

Now we also get one more solution of \(x = 0\), but if you put that value in the equation, we get LHS as \(\frac{0}{0}\), which in language of Mathematics is defined as Undetermined, meaning its a value that cannot be determined based on the current rules of Mathematics. Another such Undetermined is \(\frac{\infty}{\infty}\)

Hence, \(x = 0\) cannot be a possible solution

Hence, Answer are B and E

Re: Which of the following values of x satisfy the above equatio
[#permalink]
18 Jan 2022, 09:02

Why 0 isn't a valid solution?

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:

Cheers,

Brent

GreenlightTestPrep wrote:

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:

Show: ::

B and E

Cheers,

Brent

Re: Which of the following values of x satisfy the above equatio
[#permalink]
18 Jan 2022, 10:11

1

Expert Reply

Chaithraln2499 wrote:

Why 0 isn't a valid solution?

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 (0/0 is undefined).

So, x = 0 is NOT a valid solution.

_________________

Which of the following values of x satisfy the above equatio
[#permalink]
23 Jan 2022, 14:57

1

After we have performed the algebra nicely described in earlier posts, the question becomes:

**Quote:**

For any quadratic of the form \(ax^2 + bx + c = 0\):

The sum of the roots \(= -\frac{b}{a}\)

The product of the roots \(= \frac{c}{a}\)

Given \(2x^2+3x-27=0\):

The sum of the roots \(= -\frac{b}{a} = -\frac{3}{2} = -1.5\)

The product of the roots \(=\frac{c}{a} = \frac{-27}{2} = -13.5\)

We can PLUG IN THE ANSWERS.

A sum of -1.5 will be yielded only by the following combinations:

-6 + 4.5 = -1.5

-4.5 + 3 = -1.5

Of these two options, only the second will yield a product of -13.5:

-4.5 * 3 = -13.5

The correct answers are B and E.

_________________

Which of the following values satisfy the equation \(2x^2+3x-27=0\)?

Indicate all possible values.

A) -6

B) -4.5

C) -3

D) 0

E) 3

F) 4.5

G) 6

Indicate all possible values.

A) -6

B) -4.5

C) -3

D) 0

E) 3

F) 4.5

G) 6

For any quadratic of the form \(ax^2 + bx + c = 0\):

The sum of the roots \(= -\frac{b}{a}\)

The product of the roots \(= \frac{c}{a}\)

Given \(2x^2+3x-27=0\):

The sum of the roots \(= -\frac{b}{a} = -\frac{3}{2} = -1.5\)

The product of the roots \(=\frac{c}{a} = \frac{-27}{2} = -13.5\)

We can PLUG IN THE ANSWERS.

A sum of -1.5 will be yielded only by the following combinations:

-6 + 4.5 = -1.5

-4.5 + 3 = -1.5

Of these two options, only the second will yield a product of -13.5:

-4.5 * 3 = -13.5

The correct answers are B and E.

_________________

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