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GMATPrepNow wrote:
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
Approach #1If 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 =
27If \(x^{\frac{1}{3}}\) = -8, then x =
-8So, the SUM of the solutions =
27 +
-8 = 19
Answer: D
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Approach #2If 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 =
27If \(x^{\frac{1}{3}}\) = -8, then x =
-8So, the SUM of the solutions =
27 +
-8 = 19
Answer: D
Cheers,
Brent
method ? I didn't get it.