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16 Sep 2014, 01:23
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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. \(5\)
A. \(-2\)
B. \(0\)
C. \(1\)
D. \(3\)
E. \(5\)
16 Sep 2014, 01:23
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Bookmarks
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
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
14 Oct 2014, 11:06
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hessen923
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.
