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What is the sum of all solutions to the equation

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What is the sum of all solutions to the equation  [#permalink]

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New post 12 Oct 2018, 08:33
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8
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A
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C
D
E

Difficulty:

  45% (medium)

Question Stats:

57% (01:47) correct 43% (02:52) wrong based on 59 sessions

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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

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Re: What is the sum of all solutions to the equation  [#permalink]

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New post 12 Oct 2018, 09:29
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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


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.

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What is the sum of all solutions to the equation  [#permalink]

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New post 12 Oct 2018, 11:57
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



hey GMATPrepNow Brent :) can you please provide your detailed solution :)

have a good weekend :)
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What is the sum of all solutions to the equation  [#permalink]

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New post Updated on: 13 Oct 2018, 06:35
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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 #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
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Originally posted by GMATPrepNow on 12 Oct 2018, 12:25.
Last edited by GMATPrepNow on 13 Oct 2018, 06:35, edited 2 times in total.
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Re: What is the sum of all solutions to the equation  [#permalink]

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New post 12 Oct 2018, 12:26
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dave13 wrote:
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



hey GMATPrepNow Brent :) can you please provide your detailed solution :)

have a good weekend :)


Done!
Have a good weekend yourself!

Cheers,
Brent
_________________

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VP
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Re: What is the sum of all solutions to the equation  [#permalink]

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New post 13 Oct 2018, 00:44
GMATPrepNow wrote:
[*]
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 #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 \(x^{\frac{1}{3}}\) = -8, 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 \(x^{\frac{1}{3}}\) = -8, then x = -8

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

Answer: D

Cheers,
Brent



Hello Brent, GMATPrepNow

thank you for great explanation :) just one question: can you please explain when to use u-substitution method ? I didn't get it.

thanks!
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Re: What is the sum of all solutions to the equation  [#permalink]

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New post 13 Oct 2018, 05:13
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dave13 wrote:
Hello Brent, GMATPrepNow

thank you for great explanation :) just one question: can you please explain when to use u-substitution method ? I didn't get it.
thanks!


As you can see from my solutions above, it isn't absolutely necessary to use u-substitution here, but it does help make things clearer.
I'd say one might CONSIDER using u-substitution in cases where a particular term appears more than once.

For example: (k² + 7k - 1)² = 10(k² + 7k - 1) + 24
If we let u = (k² + 7k - 1), our equation becomes u² = 10u + 24 (much easier to deal with)

Does that help?

Cheers,
Brent
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Re: What is the sum of all solutions to the equation   [#permalink] 13 Oct 2018, 05:13
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