B,E

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

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


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

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

Bump for further discussion

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.

It took a lot of time

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

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

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

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

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

\(x^{19} = 0\) 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

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.

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



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.

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.