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D01-13 31 Jan 2025, 15:07
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I like the solution - it’s helpful.
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D01-13 01 Feb 2025, 12:45
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D01-13 12 Feb 2025, 12:51
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I kindly ask if someone can shed some light on my flawed approach. I can see my solution is mechanical and lacks creativity; nevertheless, it seems to me mathematically sound.

x^4 = x^3 + 6x^2
6x^2 - x = 0
x(6x - 1) = 0

thus, x is 0 or 1/6 :(

P.S Bunuel you are a freaking legend!
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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
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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D01-13 12 Feb 2025, 12:59
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amirezzaee
I kindly ask if someone can shed some light on my flawed approach. I can see my solution is mechanical and lacks creativity; nevertheless, it seems to me mathematically sound.

x^4 = x^3 + 6x^2
6x^2 - x = 0
x(6x - 1) = 0

thus, x is 0 or 1/6 :(

P.S Bunuel you are a freaking legend!
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The red part is incorrect: x^3 - x^4 is not equal to -x. It looks like you mistakenly treated it as if it were 3x - 4x.
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D01-13 20 Feb 2025, 23:23
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I don’t quite agree with the solution. 4th root of 16 can be -2 (-2*-2*-2*-2 =16), -2 is a valid solution
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D01-13 20 Feb 2025, 23:42
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soumya151299
I don’t quite agree with the solution. 4th root of 16 can be -2 (-2*-2*-2*-2 =16), -2 is a valid solution
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Your thinking is not correct.


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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D01-13 13 Jun 2025, 09:17
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I don’t quite agree with the solution. If x takes negative value, the expression inside the 4th root sign doesn't become negative hence -2 should not be rejected
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D01-13 22 Jun 2025, 11:34
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I don’t quite agree with the solution. If x takes negative value, the expression inside the 4th root sign doesn't become negative hence -2 should not be rejected
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x = -2 is not rejected because the expression under the square root becomes negative. It is rejected because in this case we get \(-2 = \sqrt[4]{x^3 + 6x^2}\), which cannot be true since even roots cannot produce negative results. Please review the entire discussion carefully again.
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