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12 Oct 2018, 09:33
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
69% (02:00) correct
31%
(02:37)
wrong
based on 540
sessions
History
Date
Time
Result
Not Attempted Yet
What is the sum of all solutions to the equation \(x^{\frac{2}{3}} - x^{\frac{1}{3}} - 2 = 4\)?
A) -35
B) -19
C) 7
D) 19
E) 35
A) -35
B) -19
C) 7
D) 19
E) 35
Updated on: 13 Oct 2018, 07:35
Originally posted by BrentGMATPrepNow on 12 Oct 2018, 13:25.
Last edited by BrentGMATPrepNow on 13 Oct 2018, 07:35, edited 2 times in total.
Last edited by BrentGMATPrepNow on 13 Oct 2018, 07:35, edited 2 times in total.
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[*]
Approach #1
If we recognize that \(x^{\frac{2}{3}} = (x^{\frac{1}{3}})^2\), then we can use a technique called u-substitution.
Let \(u = x^{\frac{1}{3}}\)
Now take original equation and replace \(x^{\frac{1}{3}}\) 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 \(x^{\frac{1}{3}}\) to get:
\(x^{\frac{1}{3}}\) = 3 and \(x^{\frac{1}{3}}\) = -2
If \(x^{\frac{1}{3}}\) = 3, then x = 27
If \(x^{\frac{1}{3}}\) = -2, then x = -8
So, the SUM of the solutions = 27 + -8 = 19
Answer: D
---------------------------------------------------------
Approach #2
If we recognize that \(x^{\frac{2}{3}} = (x^{\frac{1}{3}})^2\), then we can go straight to factoring.
GIVEN: \(x^{\frac{2}{3}} - x^{\frac{1}{3}} - 2 = 4\)
Subtract 4 from both sides to get: \(x^{\frac{2}{3}} - x^{\frac{1}{3}} - 6 = 0\)
Factor: \((x^{\frac{1}{3}} - 3)(x^{\frac{1}{3}} + 2) = 0\)
So, EITHER \(x^{\frac{1}{3}}\) = 3 OR \(x^{\frac{1}{3}}\) = -2
If \(x^{\frac{1}{3}}\) = 3, then x = 27
If \(x^{\frac{1}{3}}\) = -2, then x = -8
So, the SUM of the solutions = 27 + -8 = 19
Answer: D
Cheers,
Brent
GMATPrepNow
Approach #1
If we recognize that \(x^{\frac{2}{3}} = (x^{\frac{1}{3}})^2\), then we can use a technique called u-substitution.
Let \(u = x^{\frac{1}{3}}\)
Now take original equation and replace \(x^{\frac{1}{3}}\) 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 \(x^{\frac{1}{3}}\) to get:
\(x^{\frac{1}{3}}\) = 3 and \(x^{\frac{1}{3}}\) = -2
If \(x^{\frac{1}{3}}\) = 3, then x = 27
If \(x^{\frac{1}{3}}\) = -2, then x = -8
So, the SUM of the solutions = 27 + -8 = 19
Answer: D
---------------------------------------------------------
Approach #2
If we recognize that \(x^{\frac{2}{3}} = (x^{\frac{1}{3}})^2\), then we can go straight to factoring.
GIVEN: \(x^{\frac{2}{3}} - x^{\frac{1}{3}} - 2 = 4\)
Subtract 4 from both sides to get: \(x^{\frac{2}{3}} - x^{\frac{1}{3}} - 6 = 0\)
Factor: \((x^{\frac{1}{3}} - 3)(x^{\frac{1}{3}} + 2) = 0\)
So, EITHER \(x^{\frac{1}{3}}\) = 3 OR \(x^{\frac{1}{3}}\) = -2
If \(x^{\frac{1}{3}}\) = 3, then x = 27
If \(x^{\frac{1}{3}}\) = -2, then x = -8
So, the SUM of the solutions = 27 + -8 = 19
Answer: D
Cheers,
Brent
General Discussion
12 Oct 2018, 10:29
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GMATPrepNow
Let \(x^{\frac{1}{3}} =a\)
So the equation can be written as, a^2 - a - 6=0
Roots of above equation are 3 and -2.
So, \(x^{\frac{1}{3}} = 3\) or x = 27.
\(x^{\frac{1}{3}} = -2\) or x = -8.
The sum of the roots are 27-8=19.
Hence D.
Cheers!
can you please provide your detailed solution 
