M25-13

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. \(5\)
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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
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Hi Bunuel,

For the expression under the root to be valid
x^3+6*x^2 >=0
=> X^2(x+6)>=0

x^2 will always be greater than or equal to zero. The required condition is (x+6)>=0 => x>=(-6)

This condition is satisfied by all the solutions of the given equation. Hence, the answer should be 0+3-2 = 1.

Please let me know where I'm going wrong.

TIA

You have to read solution carefully: \(x\) 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 \(x=3\).
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Hi Bunuel,

I reached to solution by following way.

x^4=x^3+6x^2
X^4=X^2(X+6)
(X^4)/X^2=X+6
X^2=X+6
X^2-X=6
X(X-1)=6
X(X-1)=2*3
X=3, X=3

Is above approach wrong.

Regards,
Ammu

Yes, it's wrong. What does the red part even mean?

Also, if you divide (reduce) x^4 = x^2(x + 6) by x^2 you assume, with no ground for it, that x does not equal to zero thus exclude a possible solution (notice that x=0 satisfies the equation). Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

Correct way of solving this question is given above.
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If x = -2, then what's inside the fourth-root sign is 16. Why exactly can't the fourth root of 16 be -2?

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.
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Thank you for the explanation. However, I am still puzzled.

If x^2 = 9, then x = √9. So why does the first equation have two solutions, while the second equation only one?

Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).
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I got your point. Thank you for explanation and suggestion (Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero.)

Thanks, but its 'non-intuitive' while doing the calculation. Any suggestions to keep this in mind while being in the 'flow'. This is perhaps the only odd thing on the GMAT quant!

Practice should help.


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i think its a bit confusing because if u divide all sides by x^2 the solution would be different

Please elaborate what you mean. Thank you.
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Hi Bunuel,

Had the question been in the form x^4 = x^3 + 6*x^2, then x would have 3 feasible solutions (0, 3, -2). Correct me if I am wrong. As the stated question has 4th root and we had to clear the root hence x = -2 would not be a feasible solution.

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Yes, that's correct.
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Bring the 4th root to LHS..

x^4=x^3+6x^2

x^4 - x^3 - 6x^2 = 0

Common taking out

x^2(x^2 - x - 6) = 0

We know that if x^2 = 0, we have two roots as 0,0

Now x^2 - x - 6 = 0

we get x = 3, -2

But we have 4th root for x.. so x cannot be negative.. Hence -2 is ruled out.. Now sum of the roots.. 0 + 0 + 3 = 3

Note: if you are confused with how -2 cannot be the root.. Put -2 in the main equation.. x = 4th root (x^3 + 6 x^2) => -2 = 4th root (16) => -2 = 2.. Hence ruled out..

Hi Bunuel ,

I am little confused here :
√√ (x³ + 6x²)
For x = -2
expression (x³ + 6x²) = -8 +24 = 16 a positive number

We should not ignore x = -2 here
any thing under roots should be +ve in GMAT - hence here 16 is positive for x = -2
we do not have only x as under root but an expression.

Thanks,
Divya

You are confused because you did not read the question, solution and discussion above carefully.

Given: \(x=\sqrt[4]{x^3+6x^2}\) --> x is equal to the 4th root of some number, thus x CANNOT be negative.
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