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D01-13

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Bunuel
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D01-13 [#permalink] New post   16 Sep 2014, 00:11
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If \(x = \sqrt[4]{x^3 + 6x^2}\), what is the sum of all possible values of \(x\)?

A. \(-2\)
B. \(0\)
C. \(1\)
D. \(3\)
E. \(5\)
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D
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Bunuel
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Re D01-13 [#permalink] New post   16 Sep 2014, 00:11
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Official Solution:

If \(x = \sqrt[4]{x^3 + 6x^2}\), what is the sum of all possible values of \(x\)?

A. \(-2\)
B. \(0\)
C. \(1\)
D. \(3\)
E. \(5\)


Begin by raising the given expression to the 4th power: \(x^4 = x^3 + 6x^2\);

Next, rearrange the equation and factor out \(x^2\): \(x^2(x^2 - x - 6) = 0\);

Now, factorize the expression: \(x^2(x - 3)(x + 2) = 0\);

This results in three roots: \(x = 0\), \(x = 3\), and \(x = -2\). However, \(x\) cannot be negative since it is equal to the 4th root of an expression (\(\sqrt[4]{\text{expression}} \geq 0\)), which is always non-negative. Therefore, only two solutions 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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D01-13 [#permalink] New post   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, \({-2}^4 = -16\), not -16. It would be -16 if it were written as \({(-2)}^4\).

Next, when the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), 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] New post   02 Oct 2014, 07:12
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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] New post   23 Oct 2014, 04:00
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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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Re: D01-13 [#permalink] New post   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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D01-13 [#permalink] New post   01 Nov 2014, 04:51
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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.


When dividing the equation x^4 = x^2(x + 6) by x^2, it is important to note that this assumes x ≠ 0, without any basis for this assumption. This excludes a potential solution, as x = 0 does indeed satisfy the equation. It is crucial to remember that you should never simplify an equation by dividing it by a variable (or an expression containing a variable) if you cannot guarantee that the variable (or the expression with the variable) is not equal to zero. Dividing by zero is not permitted.

Check more tips on Algebra here: https://gmatclub.com/forum/algebra-tips ... 75003.html

Hope it helps.
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D01-13 [#permalink] New post   01 Apr 2017, 09:38
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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.
Answer: 3+0 = 3
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Re: D01-13 [#permalink] New post   29 Jun 2018, 01:23
I think this is a poor-quality question and I don't agree with the explanation. the solution of minus two is correct , you can have the 4th root of (X=-2) it is a 4th root of 16 ,and 16 it is a positive number which can use for the root.
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D01-13 [#permalink] New post   29 Jun 2018, 01:30
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Eyaltau wrote:
I think this is a poor-quality question and I don't agree with the explanation. the solution of minus two is correct , you can have the 4th root of (X=-2) it is a 4th root of 16 ,and 16 it is a positive number which can use for the root.


\(\sqrt{}\) denotes a function. Mathematically the square root function cannot give negative result. 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.

OFFICIAL GUIDE:
\(\sqrt{n}\) denotes the positive number whose square is n.
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Re: D01-13 [#permalink] New post   25 Apr 2020, 07:53
I think this is a high-quality question and I agree with explanation.
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Re: D01-13 [#permalink] New post   08 Oct 2020, 09:18
I think this is a high-quality question and I don't agree with the explanation. Why 4th root or even root of some no can't be -ve? Any even root of a no can +ve as well as -ve
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D01-13 [#permalink] New post   08 Oct 2020, 09:23
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sandipbhuyan wrote:
I think this is a high-quality question and I don't agree with the explanation. Why 4th root or even root of some no can't be -ve? Any even root of a no can +ve as well as -ve


I'll try once more.

Even roots cannot give negative result.

\(\sqrt{...}\) is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign (\(\sqrt{...}\)) always means non-negative square root.


The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;
Similarly \(\sqrt{\frac{1}{16}} = \frac{1}{4}\), NOT +1/4 or -1/4.


Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).
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Re: D01-13 [#permalink] New post   24 Nov 2020, 05:09
I think this is a high-quality question and I agree with explanation.
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Re: D01-13 [#permalink] New post   27 Mar 2021, 12:40
I think this is a high-quality question and I agree with explanation.
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Re: D01-13 [#permalink] New post   29 Apr 2023, 11:07
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re D01-13 [#permalink] New post   30 Jul 2023, 05:55
I think this is a high-quality question and I agree with explanation.
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Re D01-13 [#permalink] New post   12 Aug 2023, 09:32
I think this is a high-quality question and I agree with explanation.
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