GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Jan 2019, 08:54

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• ### The winners of the GMAT game show

January 22, 2019

January 22, 2019

10:00 PM PST

11:00 PM PST

In case you didn’t notice, we recently held the 1st ever GMAT game show and it was awesome! See who won a full GMAT course, and register to the next one.
• ### Key Strategies to Master GMAT SC

January 26, 2019

January 26, 2019

07:00 AM PST

09:00 AM PST

Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.

# D01-13

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 52386

### Show Tags

15 Sep 2014, 23:11
4
27
00:00

Difficulty:

95% (hard)

Question Stats:

35% (01:30) correct 65% (01:03) wrong based on 339 sessions

### HideShow timer Statistics

If $$x=\sqrt[4]{x^3+6x^2}$$, then the sum of all possible solutions for x is:

A. $$-2$$
B. $$0$$
C. $$1$$
D. $$3$$
E. $$5$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 52386

### Show Tags

15 Sep 2014, 23:11
6
10
Official Solution:

If $$x=\sqrt[4]{x^3+6x^2}$$, then the sum of all possible solutions for x is:

A. $$-2$$
B. $$0$$
C. $$1$$
D. $$3$$
E. $$5$$

Take the given expression to the 4th power: $$x^4=x^3+6x^2$$;

Re-arrange and factor out x^2: $$x^2(x^2-x-6)=0$$;

Factorize: $$x^2(x-3)(x+2)=0$$;

So, the roots are $$x=0$$, $$x=3$$ and $$x=-2$$. But $$x$$ cannot be negative as it equals to the even (4th) root of some expression ($$\sqrt{expression}\geq{0}$$), thus only two solution are valid $$x=0$$ and $$x=3$$.

The sum of all possible solutions for x is 0+3=3.

_________________
Manager
Joined: 29 Apr 2014
Posts: 67
Location: India
Concentration: Other, Finance
GMAT 1: 770 Q51 V42
WE: Analyst (Retail Banking)

### Show Tags

02 Oct 2014, 06:12
2
Bunuel wrote:
Official Solution:

If $$x=\sqrt[4]{x^3+6x^2}$$, then the sum of all possible solutions for x is:

A. $$-2$$
B. $$0$$
C. $$1$$
D. $$3$$
E. $$5$$

Take the given expression to the 4th power: $$x^4=x^3+6x^2$$;

Re-arrange and factor out x^2: $$x^2(x^2-x-6)=0$$;

Factorize: $$x^2(x-3)(x+2)=0$$;

So, the roots are $$x=0$$, $$x=3$$ and $$x=-2$$. But $$x$$ cannot be negative as it equals to the even (4th) root of some expression ($$\sqrt{expression}\geq{0}$$), thus only two solution are valid $$x=0$$ and $$x=3$$.

The sum of all possible solutions for x is 0+3=3.

Why can 4th root of an expression not be negative?

$${-2}^4 = 16$$ and $$2^4 = 16$$

So, shouldn't $$\sqrt[4]{16}$$ be $$-2$$ or $$2$$?
_________________

Psst...If that helped press the thumb on the left

Math Expert
Joined: 02 Sep 2009
Posts: 52386

### Show Tags

02 Oct 2014, 14:16
5
2
eshan429 wrote:
Bunuel wrote:
Official Solution:

If $$x=\sqrt[4]{x^3+6x^2}$$, then the sum of all possible solutions for x is:

A. $$-2$$
B. $$0$$
C. $$1$$
D. $$3$$
E. $$5$$

Take the given expression to the 4th power: $$x^4=x^3+6x^2$$;

Re-arrange and factor out x^2: $$x^2(x^2-x-6)=0$$;

Factorize: $$x^2(x-3)(x+2)=0$$;

So, the roots are $$x=0$$, $$x=3$$ and $$x=-2$$. But $$x$$ cannot be negative as it equals to the even (4th) root of some expression ($$\sqrt{expression}\geq{0}$$), thus only two solution are valid $$x=0$$ and $$x=3$$.

The sum of all possible solutions for x is 0+3=3.

Why can 4th root of an expression not be negative?

$${-2}^4 = 16$$ and $$2^4 = 16$$

So, shouldn't $$\sqrt[4]{16}$$ be $$-2$$ or $$2$$?

First of all $${-2}^4 = -16$$, it would be 16 if it were $${(-2)}^4$$.

Next, when the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.

That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.
_________________
Manager
Joined: 23 Jan 2013
Posts: 138
Concentration: Technology, Other
Schools: Berkeley Haas
GMAT Date: 01-14-2015
WE: Information Technology (Computer Software)

### Show Tags

17 Oct 2014, 21:44
Bunuel wrote:
Official Solution:

If $$x=\sqrt[4]{x^3+6x^2}$$, then the sum of all possible solutions for x is:

A. $$-2$$
B. $$0$$
C. $$1$$
D. $$3$$
E. $$5$$

Take the given expression to the 4th power: $$x^4=x^3+6x^2$$;

Re-arrange and factor out x^2: $$x^2(x^2-x-6)=0$$;

Factorize: $$x^2(x-3)(x+2)=0$$;

[highlight]So, the roots are $$x=0$$, $$x=3$$ and $$x=-2$$. But $$x$$ cannot be negative as it equals to the even (4th) root of some expression ($$\sqrt{expression}\geq{0}$$), thus only two solution are valid $$x=0$$ and $$x=3$$.[/highlight]

The sum of all possible solutions for x is 0+3=3.

Hi Bunnel ,

Could you elaborate more on the above statement ?
If i substitute X = 0 , -2 , 3 Equation turns out to be 0 in all cases which it statisfies hence i feel all answers are correct . Maybe my thinking is flawed .

Thanks and Regards ,
Shelrod007
Math Expert
Joined: 02 Sep 2009
Posts: 52386

### Show Tags

18 Oct 2014, 01:11
shelrod007 wrote:
Bunuel wrote:
Official Solution:

If $$x=\sqrt[4]{x^3+6x^2}$$, then the sum of all possible solutions for x is:

A. $$-2$$
B. $$0$$
C. $$1$$
D. $$3$$
E. $$5$$

Take the given expression to the 4th power: $$x^4=x^3+6x^2$$;

Re-arrange and factor out x^2: $$x^2(x^2-x-6)=0$$;

Factorize: $$x^2(x-3)(x+2)=0$$;

[highlight]So, the roots are $$x=0$$, $$x=3$$ and $$x=-2$$. But $$x$$ cannot be negative as it equals to the even (4th) root of some expression ($$\sqrt{expression}\geq{0}$$), thus only two solution are valid $$x=0$$ and $$x=3$$.[/highlight]

The sum of all possible solutions for x is 0+3=3.

Hi Bunnel ,

Could you elaborate more on the above statement ?
If i substitute X = 0 , -2 , 3 Equation turns out to be 0 in all cases which it statisfies hence i feel all answers are correct . Maybe my thinking is flawed .

Thanks and Regards ,
Shelrod007

Yes, your are not right. Have you read the whole thread? How does -2 satisfy $$x=\sqrt[4]{x^3+6x^2}$$?
_________________
Manager
Joined: 02 Jul 2012
Posts: 187
Location: India
Schools: IIMC (A)
GMAT 1: 720 Q50 V38
GPA: 2.6
WE: Information Technology (Consulting)

### Show Tags

23 Oct 2014, 03:00
1
Bunuel wrote:
If $$x=\sqrt[4]{x^3+6x^2}$$, then the sum of all possible solutions for x is:

A. $$-2$$
B. $$0$$
C. $$1$$
D. $$3$$
E. $$5$$

I got this question wrong in the test. So, solving it here for a better understanding.

Basically this question tests one concept i.e. $$\sqrt{x^2}$$ is x and not -x.

Solving the equation, $$x^4= x^3+6x^2$$
$$=>$$ $$x^4-x^3-6x^2=0$$
$$x^2(x^2-x-6)=0$$
$$x=0,3,-2$$

As mentioned earlier, $$\sqrt{x^2}$$ is x and not -x, the value -2 is discarded.

Thus the two values are 3 and 0

SO, the sum = 3
_________________

Give KUDOS if the post helps you...

Intern
Joined: 26 Jun 2014
Posts: 1

### Show Tags

31 Oct 2014, 15:26
would it be wrong to have $$x^4 = x^3+6x^2$$ (as previously stated)
then factor out $$x^2$$ on both sides for$$(x^2)(x^2) = (x^2)(x+6)$$
then divide both sides by $$x^2$$ leaving $$x^2=x+6$$, ie. $$x^2-x-6=0$$?

if this is acceptable does that still leave 0 as a solution with -2 and 3?
I'm more curious if this would cause problems on other similar problems.
Manager
Joined: 02 Jul 2012
Posts: 187
Location: India
Schools: IIMC (A)
GMAT 1: 720 Q50 V38
GPA: 2.6
WE: Information Technology (Consulting)

### Show Tags

31 Oct 2014, 19:59
SQUEE wrote:
would it be wrong to have $$x^4 = x^3+6x^2$$ (as previously stated)
then factor out $$x^2$$ on both sides for$$(x^2)(x^2) = (x^2)(x+6)$$
then divide both sides by $$x^2$$ leaving $$x^2=x+6$$, ie. $$x^2-x-6=0$$?

if this is acceptable does that still leave 0 as a solution with -2 and 3?
I'm more curious if this would cause problems on other similar problems.

Dear SQUEE,
If you would divide both sides by x^2 this would even contain the possibility of dividing by 0.
As we know that we can not divide by 0 since division by 0 is not defined. So do dividing by x^2 would be wrong.

Hope that helps you.
_________________

Give KUDOS if the post helps you...

Math Expert
Joined: 02 Sep 2009
Posts: 52386

### Show Tags

01 Nov 2014, 03:51
SQUEE wrote:
would it be wrong to have $$x^4 = x^3+6x^2$$ (as previously stated)
then factor out $$x^2$$ on both sides for$$(x^2)(x^2) = (x^2)(x+6)$$
then divide both sides by $$x^2$$ leaving $$x^2=x+6$$, ie. $$x^2-x-6=0$$?

if this is acceptable does that still leave 0 as a solution with -2 and 3?
I'm more curious if this would cause problems on other similar problems.

Also, if you divide (reduce) x^4 = x^2(x + 6) by x^2 you assume, with no ground for it, that x does not equal to zero thus exclude a possible solution (notice that x=0 satisfies the equation). Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.

Check more tips on Algebra here: algebra-tips-and-hints-175003.html

Hope it helps.
_________________
Manager
Joined: 30 Mar 2013
Posts: 109

### Show Tags

08 Dec 2014, 12:31
if we take x=-2, the solving the equation, LHS would have -2, and the RHS would have a positive 2. Therefore, they are not equal and -2 doesn't satisfy the equation. Plugging the numbers back in help.
Intern
Joined: 10 Dec 2014
Posts: 29
GMAT Date: 12-30-2014

### Show Tags

14 Dec 2014, 21:16
We need not even consider -2 and check it.Because the root will never give a negative value
Current Student
Joined: 15 Apr 2015
Posts: 13
Location: India
Schools: EDHEC (A\$)
GMAT 1: 620 Q44 V31
GPA: 3.25

### Show Tags

10 Sep 2015, 06:19
I don't agree with the explanation. i don't agree with the explanation of why you have chosen to not include -2 in the sum.
Also after taking the 4th power of the equation why can't we use the general formula for n powered equation where sum of roots =-b/a.
i got 1 using that formula
Math Expert
Joined: 02 Sep 2009
Posts: 52386

### Show Tags

10 Sep 2015, 07:19
mrigoel wrote:
I don't agree with the explanation. i don't agree with the explanation of why you have chosen to not include -2 in the sum.
Also after taking the 4th power of the equation why can't we use the general formula for n powered equation where sum of roots =-b/a.
i got 1 using that formula

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, erc. then the only accepted answer is the positive root.

That is:
$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt[4]{16} = 2$$, NOT +2 or -2;

Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and -3.

Hope it's clear.
_________________
Intern
Joined: 05 Aug 2015
Posts: 42

### Show Tags

02 Nov 2015, 08:36
I got x=0, 3, -2 but failed to check to see if all the solutions are valid. Do I always need to check if the roots are valid or just on certain types of problems? If so, which types / under what conditions?

Just want to better understand and save time on the test. Thanks so much Gmatclub!
_________________

Working towards 25 Kudos for the Gmatclub Exams - help meee I'm poooor

Manager
Joined: 27 Aug 2014
Posts: 74

### Show Tags

26 Feb 2016, 06:20
Bunuel wrote:
If $$x=\sqrt[4]{x^3+6x^2}$$, then the sum of all possible solutions for x is:

A. $$-2$$
B. $$0$$
C. $$1$$
D. $$3$$
E. $$5$$

Hi

Also my little bit. When reducing, we hit a quadratic equation x^2-x-6=0 that should have only two roots. So that should give some hint that all three roots cant be acceptable.
Intern
Joined: 07 Dec 2015
Posts: 2
Location: Nepal
Concentration: Finance, Statistics
GPA: 3

### Show Tags

17 Mar 2016, 07:24
I think this the explanation isn't clear enough, please elaborate. i dont understand why -2 is not a solution .
for instance squrt of 4 has 2 soln i.e 2 and -2. please explain?
Math Expert
Joined: 02 Sep 2009
Posts: 52386

### Show Tags

17 Mar 2016, 09:27
2
1
I think this the explanation isn't clear enough, please elaborate. i dont understand why -2 is not a solution .
for instance squrt of 4 has 2 soln i.e 2 and -2. please explain?

That's not true.

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.

That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.
_________________
Intern
Joined: 11 Jan 2015
Posts: 34

### Show Tags

26 Mar 2016, 15:24
Hi Bunuel, would the following be a valid approach to this question or just luck?

1.) Re-arrange and factor out $$x^2: x^2(x^2−x−6)=0$$ --> So one solution is 0 -> $$x^2$$
2.) Use discriminant for the left equation $$(x^2−x−6)=0$$: $$b^2 - 4ac$$ --> 1-(4*1*(-6)) -> 25 -> Since 25 is positive the left equation has two solutions
3.) 1.) + 2.) = 3 Solutions
Math Expert
Joined: 02 Sep 2009
Posts: 52386

### Show Tags

27 Mar 2016, 01:47
Hi Bunuel, would the following be a valid approach to this question or just luck?

1.) Re-arrange and factor out $$x^2: x^2(x^2−x−6)=0$$ --> So one solution is 0 -> $$x^2$$
2.) Use discriminant for the left equation $$(x^2−x−6)=0$$: $$b^2 - 4ac$$ --> 1-(4*1*(-6)) -> 25 -> Since 25 is positive the left equation has two solutions
3.) 1.) + 2.) = 3 Solutions

You should find the actual roots and find out whether all of them are valid. Only 2 of them are valid for the equation at hand. Please refer to the discussion on previous 2 pages.
_________________
Re: D01-13 &nbs [#permalink] 27 Mar 2016, 01:47

Go to page    1   2   3    Next  [ 47 posts ]

Display posts from previous: Sort by

# D01-13

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.