Given $x = \sqrt[4]{x^{3}+6x^{2}}$, determine whether the sum of all possible roots of $x$ is equal to $1$. Given $$x = \sqrt[4]{x^{3}+6x^{2}}$$
Quantity $A$: Sum of all possible roots of $x$
Quantity $B$: $1$
My solution: I've taken fourth power on both sides of the equation to get $$x^{4} = x^{3} + 6x^{2}.$$
Rearranging and factorizing $x^{2}$ gives $$x^{2}(x^{2} - x - 6) = 0.$$
Further factorizing gives $$x^{2}(x-3)(x+2) = 0.$$
So the roots of $x$ are $0$,$-2$,and $3$.
So the sum of the roots is $1$. Thus, my answer is $C$ $(A=B)$.
But the answer in the practice test was $A$. Why is this so?

$-2$ is not a solution of $x = \sqrt[4]{x^{3}+6x^{2}}$.
