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What is the sum of all solutions to the

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What is the sum of all solutions to the

by Brent@GMATPrepNow » Tue Feb 28, 2017 8:33 am
What is the sum of all solutions to the equation 3x^(2/3) = 54 + 9x^(1/3)
A) 63
B) 189
C) 216
D) 243
E) 567

Answer: B

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by regor60 » Tue Feb 28, 2017 9:17 am
Rearrange the equation:

3x(2/3)-9x^(1/3)-54 = 0

Let Y = X(1/3) to simplify things

3Y^2-9Y-54 = 0 >divide through by 3

Y^2-3Y-18=0

Factor: (Y-6)*(y+3) = 0 > Y=6 or Y=-3

X=Y^3, therefore X= 216 or -27

Adding solutions together 216-27=189, B
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by Brent@GMATPrepNow » Wed Mar 01, 2017 5:49 am
Brent@GMATPrepNow wrote:What is the sum of all solutions to the equation 3x^(2/3) = 54 + 9x^(1/3)
A) 63
B) 189
C) 216
D) 243
E) 567
We should start by recognizing that this is a QUADRATIC EQUATION in disguise.

Notice that x^(2/3) = [x^(1/3)]²

So let's let x^(1/3) = k, and replace x^(1/3) with k to get: 3k² = 54 + 9k
Rearrange to get: 3k² - 9k - 54 = 0
Factor: 3(k² - 3k - 18) = 0
Factor more: 3(k - 6)(k + 3) = 0
So, the solutions are k = 6 and k = -3

Since k = x^(1/3), we can write: x^(1/3) = 6 and x^(1/3) = -3
If x^(1/3) = 6, then x = 6³ = 216
If x^(1/3) = -3, then x = (-3)³ = -27

So, the sum of all solutions = 216 + (-27) = 189

Answer: B

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by Matt@VeritasPrep » Wed Mar 01, 2017 3:05 pm
I don't think the GMAT would ask this, though, because it can be solved with a simple formula from Algebra II, Viète's Formula. (It isn't really their style to ask questions that have trivial solutions if you know slightly higher math than they expect of you.)
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by Matt@VeritasPrep » Wed Mar 01, 2017 3:08 pm
With that in mind, let me propose an alternative solution anyway.

Let z equal the cube root of x. With that, we can express the original equation as

3z² - 9z - 54 = 0

z² - 3z - 18 = 0

(z + 3) * (z - 6) = 0

So our solutions are z = -3 and z = 6. z is the cube root of x, so our x's are -3³ and 6³, and their sum = 189.
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