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Re: GRE Math Challenge #51 [#permalink]
The answer is C.
The question is sum of all possible values of x. That means 3+(-2)=1
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Re: GRE Math Challenge #51 [#permalink]
1
Answer is C

If you put -2 in place of x you get +ve value in fourth root.
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Re: GRE Math Challenge #51-Then the sum of all possible solution [#permalink]
I can't post URLs yet, but if you type "graph of 4th root of (x^3 + 6*x^2)" into Google, it will show you the graph of the function.

The graph of (x^3+6*x^2)^(1/4) shows that x can in fact be negative, the only condition that holds is that x >= -6.

This follows from the inequality x^3+6*x^2 >= 0.
So x^2*(x+6)>= 0, therefore x+6 >= 0, therefore x >= -6

Therefore, the answer should be C, as possible roots of x include 0, 3, and -2 since -2 >= 6.

Am I missing something here?
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Re: GRE Math Challenge #51-Then the sum of all possible solution [#permalink]
when we put x=-2 we get x=2 ie is + ,so c
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Re: GRE Math Challenge #51-Then the sum of all possible solution [#permalink]
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Re: GRE Math Challenge #51-Then the sum of all possible solution [#permalink]
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