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# D01-13

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Math Expert
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D01-13 [#permalink]

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16 Sep 2014, 00:11
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Question Stats:

35% (01:52) correct 65% (01:11) wrong based on 252 sessions

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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. $$5$$
[Reveal] Spoiler: OA

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Re D01-13 [#permalink]

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16 Sep 2014, 00:11
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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.

Answer: D
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Re: D01-13 [#permalink]

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02 Oct 2014, 07:12
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.

Answer: D

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 $$2$$?
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Re: D01-13 [#permalink]

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02 Oct 2014, 15:16
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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.

Answer: D

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.
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Re: D01-13 [#permalink]

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17 Oct 2014, 22:44
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.

Answer: D

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

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Re: D01-13 [#permalink]

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18 Oct 2014, 02:11
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.

Answer: D

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 $$x=\sqrt[4]{x^3+6x^2}$$?
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Re: D01-13 [#permalink]

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23 Oct 2014, 04:00
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
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D01-13 [#permalink]

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31 Oct 2014, 16:26
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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.

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Re: D01-13 [#permalink]

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31 Oct 2014, 20:59
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.
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D01-13 [#permalink]

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01 Nov 2014, 04:51
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.
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Re: D01-13 [#permalink]

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08 Dec 2014, 13:31
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.

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Re: D01-13 [#permalink]

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14 Dec 2014, 22:16
We need not even consider -2 and check it.Because the root will never give a negative value

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Re D01-13 [#permalink]

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05 Jan 2015, 11:06
I think this question is poor.
Consider sq. root of 16 = +/- 4
now if we are gong to root this again we need +4 only but
sq. root of +4 can be +/- 2 so
4th root of 16 can be +/-2

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Re: D01-13 [#permalink]

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06 Jan 2015, 02:33
sanket1991 wrote:
I think this question is poor.
Consider sq. root of 16 = +/- 4
now if we are gong to root this again we need +4 only but
sq. root of +4 can be +/- 2 so
4th root of 16 can be +/-2

That;s totally wrong. Check here: d01-183474.html#p1422497
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Re: D01-13 [#permalink]

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06 Apr 2015, 12:30
Hi Bunuel

FYI: This question is essentially the same as M25-13, I had both of them on my Error Log today after some corrected GMATclub CATs...

Greets

DarkMight

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Re: D01-13 [#permalink]

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06 Apr 2015, 12:43
DarkMight wrote:
Hi Bunuel

FYI: This question is essentially the same as M25-13, I had both of them on my Error Log today after some corrected GMATclub CATs...

Greets

DarkMight

Thank you. Will replace one of them.
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Re D01-13 [#permalink]

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10 Sep 2015, 07:19
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

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Re: D01-13 [#permalink]

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10 Sep 2015, 08:19
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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.
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Re: D01-13 [#permalink]

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02 Nov 2015, 09:36
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!
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Re: D01-13 [#permalink]

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26 Feb 2016, 07:20
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.

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Re: D01-13   [#permalink] 26 Feb 2016, 07:20

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