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Challenge question: What is the sum of all

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Challenge question: What is the sum of all

by Brent@GMATPrepNow » Fri Oct 12, 2018 8:29 am

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What is the sum of all solutions to the equation x^(2/3) - x^(1/3) - 2 = 4?

A) -35
B) -19
C) 7
D) 19
E) 35

Answer: D
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by GMATGuruNY » Fri Oct 12, 2018 9:44 am
Brent@GMATPrepNow wrote:What is the sum of all solutions to the equation x^(2/3) - x^(1/3) - 2 = 4?

A) -35
B) -19
C) 7
D) 19
E) 35
x^(2/3) - x^(1/3) - 2 = 4

x^(2/3) - x^(1/3) - 6 = 0

[x^(1/3) - 3] [x^(1/3) + 2] = 0

Solution 1:
x^(1/3) - 3 = 0
x^(1/3) = 3
x^(1/3)³ = 3³
x = 27

Solution 2:
x^(1/3) + 2 = 0
x^(1/3) = -2
x^(1/3)³ = (-2)³
x = -8

Sum of the solutions = -8 + 27 = 19.

The correct answer is D.
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by fskilnik@GMATH » Fri Oct 12, 2018 11:30 am
Brent@GMATPrepNow wrote:What is the sum of all solutions to the equation x^(2/3) - x^(1/3) - 2 = 4?

A) -35
B) -19
C) 7
D) 19
E) 35

Source: www.gmatprepnow.com
$$?\,\,\,:\,\,\,\,\,{\text{sum}}\,\,{\text{of}}\,\,{\text{roots}}\,\,{\text{of}}\,\,\,\,\root 3 \of {{x^2}} - \root 3 \of x - 6 = 0$$
$$y = \root 3 \of x \,\,\,\,\,\,\,\,\,\left[ {\,\root 3 \of {{x^2}} = {{\left( {\root 3 \of x } \right)}^2} = {y^2}\,} \right]\,$$
$${y^2} - y - 6 = 0\,\,\,\,\,\,\,\mathop \Rightarrow \limits_{{\rm{product}}\, = \, - 6}^{{\rm{sum}}\, = \,1} \,\,\,\,\,\,\,\,\left\{ \matrix{
\,3 = {y_1} = \root 3 \of {{x_1}} \,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{cubing}}} \,\,\,\,\,{x_1} = {3^3} = 27 \hfill \cr
\, - 2 = {y_2} = \root 3 \of {{x_2}} \,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{cubing}}} \,\,\,\,\,{x_2} = {\left( { - 2} \right)^3} = - 8 \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,?\,\, = \,\,27 + \left( { - 8} \right)\,\, = \,\,19$$


This solution follows the notations and rationale taught in the GMATH method.

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Fabio.
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by Brent@GMATPrepNow » Sat Oct 13, 2018 6:33 am
Brent@GMATPrepNow wrote:What is the sum of all solutions to the equation x^(2/3) - x^(1/3) - 2 = 4?

A) -35
B) -19
C) 7
D) 19
E) 35

Answer: D
Source: www.gmatprepnow.com
Difficulty level: 600 - 650
Approach #1
If we recognize that x^(2/3) = [x^(1/3)]^2, then we can use a technique called u-substitution.
Let u = x^(1/3)

Now take original equation and replace x^(1/3) with u to get: u² - u - 2 = 4
Subtract 4 from both sides to get: to get: u² - u - 6 = 0
Factor: (u - 3)(u + 2) = 0
So, the two solutions here are u = 3 and u = -2

At this point, we replace u with x^(1/3) to get:
x^(1/3) = 3 and x^(1/3) = -2

If x^(1/3) = 3, then x = 27
If x^(1/3) = -2, then x = -8

So, the SUM of the solutions = 27 + -8 = 19

Answer: D
---------------------------------------------------------

Approach #2
If we recognize that x^(2/3) = [x^(1/3)]^2, then we can go straight to factoring.

GIVEN: x^(2/3) - x^(1/3) - 2 = 4[/m]
Subtract 4 from both sides to get:x^(2/3) - x^(1/3) - 6 = 0[/m]
Factor: (x^(1/3) - 3)(x^(1/3) + 2) = 0[/m]
So, EITHER x^(1/3) = 3 OR x^(1/3) = -2

If x^(1/3) = 3, then x = 27
If x^(1/3) = -2, then x = -8

So, the SUM of the solutions = 27 + -8 = 19

Answer: D

Cheers,
Brent
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