the answer depends on what 4 means is it 4th root, or 4* something

If we consider that the expression given is the 4th root , then the answer should be C . Can someone please explain how is the answer A ?

Agree, For the 4th root, x can be solved for 3&-2, hence sum of the roots is 1, so C should be the option.

It is said sum of all possible root:
I am getting possible values 1,3
As 3 is greater than 1 hence answer is A

The question could mean two things..
(I) If it is what it is given- \(x=4\)\(\sqrt{(x^3+6x^2)}\)
Square both sides - \(x^2=16(x^3+6x^2).......16x^3-95x^2=0....x^2(16x-95)=0\).. Sum of roots = \(0+\frac{95}{16}\) so A>B
(II) If there is a typo and it meant $th root and not 4*..
so take the 4th power - \(x^4=x^3+6x^2......x^4-x^3-6x^2=0.....x^2(x-3)(x+2)=0\) Hence, the sum is 0+3-2=1
here A=B..
so C

''x2=16(x3+6x2).......16x3−95x2=0....x2(16x−95)=0''
how is the above possible , i.e. ( -95x)?

\(\(x^2=16(x^3+6x^2).......16x^3-95x^2=0....x^2(16x-95)=0\)\)

Take out x^2 from \(16x^3-95x^2=0......x^2*16x-x^2*95=x^2(16x-95)=0\)

If presented with clear algebraic information in a Quantitative Comparison, initially look to solve algebraically.

First, raise the given equation to the 4th power to eliminate the radical resulting in x4=x3+6x2.
Next, subtract x3+6x2 from each side to set the quadratic equal to 0 in order to solve.
Then, factor x2 out of each term resulting in x2(x2−x−6) = 0.
Next, factor the interior quadratic x2−x−6 resulting in the equation x2(x−3)(x+2)=0.
Thus, it would appear that the roots are x = 0, x = 3 , and x= -2.

However, be careful to not conclude that the quantities are equal because 0 + 3 + -2 = 1.
Remember that the result of an even radical applied to a negative value is an imaginary number and thus invalid on the GRE.
So, x cannot be negative, because if it were, the result would be (4
√
(−23+6(−2)2)
=−2 which would not be an acceptable result of real arithmetic.

Subsequently, there are only two valid solutions : x = 0 and x = 3, and the sum of those valid roots will be 3 + 0 = 3, and 3 > 1.

So answer is A

How to solve this sum?

I still did not get this part

4√(−23+6(−2)2) =-2 ,

(-2)^3=-8 and 6(-2)^2=24 and 24 -8=16 . 4th root of 16 can +2 or -2. Where do we get imaginary number here.

Hey,

In the stem, something is missing hence we are not getting a clear answer. Also, this will never happen in the original exam so leave this one.

Carcass here i was able to solve till the quadratic part where x =0 or x = 3 or x = -2 but not able to go further then this

We are given the equation:

$$
\(x=\sqrt[4]{x^3+6 x^2}\)
$$


Step 1: Solve the Equation $\(x=\sqrt[4]{x^3+6 x^2}\)$
First, let's eliminate the fourth root by raising both sides to the fourth power:

$$
\(x^4=x^3+6 x^2\)
$$


Step 2: Rearrange the Equation
Bring all terms to one side to form a polynomial equation:

$$
\(x^4-x^3-6 x^2=0\)
$$


Step 3: Factor the Polynomial
Factor out $\(x^2\)$ :

$$
\(x^2\left(x^2-x-6\right)=0\)
$$


This gives us:
1. $\(x^2=0 \rightarrow x=0\)$
2. $\(x^2-x-6=0\)$

Now, solve the quadratic equation $\(x^2-x-6=0\)$ :
Using the quadratic formula:

$$
\(x=\frac{1 \pm \sqrt{1+24}}{2}=\frac{1 \pm 5}{2}\)
$$


So, the solutions are:

$$
\(\begin{gathered}
x=\frac{1+5}{2}=3 \\
x=\frac{1-5}{2}=-2
\end{gathered}\)
$$


Step 4: List All Possible Roots
From the factoring, the possible roots are:
1. $\(x=0\left(\right.\)$ from $\(\left.x^2=0\right)\)$
2. $\(x=3\)$ (from the quadratic)
3. $\(x=-2\)$ (from the quadratic)


Step 5: Verify the Roots in the Original Equation
It's crucial to verify each potential root in the original equation because raising both sides to a power can introduce extraneous solutions.
1. For $\(x=0\)$ :

$$
\(0=\sqrt[4]{0^3+6 \cdot 0^2}=\sqrt[4]{0}=0\)
$$


This holds true.
2. For $\(x=3\)$ :

$$
\(3=\sqrt[4]{3^3+6 \cdot 3^2}=\sqrt[4]{27+54}=\sqrt[4]{81}=3\)
$$


This holds true.
3. For $\(x=-2\)$ :

$$
\(-2=\sqrt[4]{(-2)^3+6 \cdot(-2)^2}=\sqrt[4]{-8+24}=\sqrt[4]{16}=2\)
$$


Here, $\(-2 \neq 2\)$, so $\(x=-2\)$ is not a valid solution.

Step 6: Identify Valid Real Roots

From the verification:
- Valid roots: $\(x=0, x=3\)$
- Invalid root: $\(x=-2\)$

Step 7: Calculate the Sum of All Possible Real Roots
Sum of valid roots:

$$
\(0+3=3\)
$$

In the end

A=3
B=1

A is greater and it is the final solution.

PS: notice how in the stem we do not have 4 before the root but is the same root of 4. I fixed it

Updated and posted a cogent solution for further discussions