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# D01-13

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Joined: 02 Sep 2009
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16 Sep 2014, 00:11
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95% (hard)

Question Stats:

31% (01:19) correct 69% (01:38) wrong based on 234 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. $$5$$

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16 Sep 2014, 00:11
6
12
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.

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02 Oct 2014, 07:12
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 $$2$$?
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02 Oct 2014, 15:16
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.
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23 Oct 2014, 04:00
2
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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31 Oct 2014, 16:26
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.
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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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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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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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10 Sep 2015, 08:19
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.
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17 Mar 2016, 08:24
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?
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17 Mar 2016, 10:27
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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.
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26 Mar 2016, 16:24
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
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27 Mar 2016, 02:47
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.
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21 Sep 2016, 10:38
how did the following question become x*4-x*3-6x*2=0?
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22 Sep 2016, 00:59
manchitkapoor wrote:
how did the following question become x*4-x*3-6x*2=0?

It does not.

Take $$x=\sqrt[4]{x^3+6x^2}$$ to the 4th power to get $$x^4=x^3+6x^2$$, which when re-arranged becomes $$x^4-x^3-6x^2=0$$. It's x to the 4th power, minus x cubed, minus 6 times x squared, equal 0.
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01 Apr 2017, 09:38
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Bunuel wrote:
manchitkapoor wrote:
how did the following question become x*4-x*3-6x*2=0?

It does not.

Take $$x=\sqrt[4]{x^3+6x^2}$$ to the 4th power to get $$x^4=x^3+6x^2$$, which when re-arranged becomes $$x^4-x^3-6x^2=0$$. It's x to the 4th power, minus x cubed, minus 6 times x squared, equal 0.

I think this should clarifies everybody's confusion regarding the answer, i hope everyone understands how we got 0,3,(-2)

$$x=\sqrt[4]{x^3+6x^2}$$
x=0
$$0=\sqrt[4]{0^3+6*0^2}$$
$$0=\sqrt[4]{0}$$
$$0=0$$

$$x=\sqrt[4]{x^3+6x^2}$$
x=3
$$3=\sqrt[4]{3^3+6*3^2}$$
$$3=\sqrt[4]{27+6*9}$$
$$3=\sqrt[4]{27+54}$$
$$3=\sqrt[4]{81}$$
$$3=3$$

$$x=\sqrt[4]{x^3+6x^2}$$
x=(-2)
$$(-2)=\sqrt[4]{(-2)^3+6*(-2)^2}$$
$$(-2)=\sqrt[4]{-8+6*4}$$
$$(-2)=\sqrt[4]{-8+24}$$
$$(-2)=\sqrt[4]{16}$$
$$(-2) =| 2$$
Thus solution x = (-2) isn't applicable.
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12 Apr 2017, 04:49
And here we go again. Sometimes i feel as if GMAT has it's own version of math that goes against universal one. I understand that on verbal there can be a whole lot of inerprteations of rules, simply because it is VERBAL, as in not a precise science which is basically a human imagination. But math has to be precise. And now, according to e-GMAT -b^m if m is even --> positive number and you are stating that −2^4=−16 and i made a whole lot of mistakes when practicing quant from both OG and scholaranium that involved sqr.root of a number, because i didn't include the negative value as well. Can comeone give a link or a screenshot of an OG question where when we take sqr root we don't consider the negative value?
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12 Apr 2017, 04:58
stan3544 wrote:
And here we go again. Sometimes i feel as if GMAT has it's own version of math that goes against universal one. I understand that on verbal there can be a whole lot of inerprteations of rules, simply because it is VERBAL, as in not a precise science which is basically a human imagination. But math has to be precise. And now, according to e-GMAT -b^m if m is even --> positive number and you are stating that −2^4=−16 and i made a whole lot of mistakes when practicing quant from both OG and scholaranium that involved sqr.root of a number, because i didn't include the negative value as well. Can comeone give a link or a screenshot of an OG question where when we take sqr root we don't consider the negative value?

Not sure I understand what you mean but GMAT doe not have its own math. That's simply not correct. The only thing is that GMAT deals with only Real Numbers: Integers, Fractions and Irrational Numbers, so no imaginary numbers. That's why the even roots from negative numbers are not defined for the GMAT.

(-2)^4 = 16 but -2^4 = -16, this is true generally not only for the GMAT.
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12 Apr 2017, 05:58
Bunuel wrote:
stan3544 wrote:
And here we go again. Sometimes i feel as if GMAT has it's own version of math that goes against universal one. I understand that on verbal there can be a whole lot of inerprteations of rules, simply because it is VERBAL, as in not a precise science which is basically a human imagination. But math has to be precise. And now, according to e-GMAT -b^m if m is even --> positive number and you are stating that −2^4=−16 and i made a whole lot of mistakes when practicing quant from both OG and scholaranium that involved sqr.root of a number, because i didn't include the negative value as well. Can comeone give a link or a screenshot of an OG question where when we take sqr root we don't consider the negative value?

Not sure I understand what you mean but GMAT doe not have its own math. That's simply not correct. The only thing is that GMAT deals with only Real Numbers: Integers, Fractions and Irrational Numbers, so no imaginary numbers. That's why the even roots from negative numbers are not defined for the GMAT.

(-2)^4 = 16 but -2^4 = -16, this is true generally not only for the GMAT.

Hey Bunuel,

Now i am totally confused. Yes it is obvious and universally understood that sqr.root of negative number is not a real number, basically it doesn't exist, why? Because no negative number raised to an even power will ever give us a negative result, yet according to your example -2^4 = -16, so we know that the number -2 raised to the power 4 gives us -16, isn't it a paradox? Now, according to the exponent rules: (-2)^4 = (-2^1)^4 = -2*(-2)*(-2)*(-2) right? Now according to your interpretation -2^4 is not the same, so it basically means -2*2*2*2, which doesn't make sense, because this goes against the whole idea behind exponents, which is: we take a number and multiply by ITSELF n times. So the correct way to write it would be -(2^4) = -16 or -(-2^4) = -16. Now back to the issue at hand: logically 2^2=4 as well as -2*(-2)=4, which is (-2)^2= 4, now i don't see why sqr.root of 4 can only be 2 and not -2 as well.
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# D01-13

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