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# What is the sum of all solutions to the equation

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CEO
Joined: 11 Sep 2015
Posts: 3447
What is the sum of all solutions to the equation  [#permalink]

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12 Oct 2018, 08:33
1
Top Contributor
8
00:00

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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Joined: 02 Aug 2015
Posts: 152
Re: What is the sum of all solutions to the equation  [#permalink]

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12 Oct 2018, 09:29
2
1
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.

Cheers!
VP
Joined: 09 Mar 2016
Posts: 1284
What is the sum of all solutions to the equation  [#permalink]

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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
CEO
Joined: 11 Sep 2015
Posts: 3447
What is the sum of all solutions to the equation  [#permalink]

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Updated on: 13 Oct 2018, 06:35
2
Top Contributor
1
[*]
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

---------------------------------------------------------

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

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.
CEO
Joined: 11 Sep 2015
Posts: 3447
Re: What is the sum of all solutions to the equation  [#permalink]

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12 Oct 2018, 12:26
1
Top Contributor
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
_________________

Test confidently with gmatprepnow.com

VP
Joined: 09 Mar 2016
Posts: 1284
Re: What is the sum of all solutions to the equation  [#permalink]

### Show Tags

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

---------------------------------------------------------

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

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!
CEO
Joined: 11 Sep 2015
Posts: 3447
Re: What is the sum of all solutions to the equation  [#permalink]

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13 Oct 2018, 05:13
Top Contributor
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
_________________

Test confidently with gmatprepnow.com

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

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