Given $x = \sqrt[4]{x^{3}+6x^{2}}$, determine whether the sum of all possible roots of $x$ is equal to $1$.

Question

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?

Answer 1

Hint. Any root $x$ of your original equation
$$x = \sqrt[4]{x^{3}+6x^{2}}$$ should be non-negative because the right-hand side is a non-negative function.

P.S. Notice that by taking fourth power on both sides you found the roots also of the equation $$-x = \sqrt[4]{x^{3}+6x^{2}}.$$

Answer 2

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