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16 Sep 2014, 00:11
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D
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Difficulty:
85%
(hard)
Question Stats:
40% (01:35) correct
60%
(01:36)
wrong
based on 491
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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. \(5\)
A. \(-2\)
B. \(0\)
C. \(1\)
D. \(3\)
E. \(5\)
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
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
02 Oct 2014, 15:16
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eshan429
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.
