Amount of solutions to an equation

Briliant.org asks for how many solutions there are to this equation:

$x^3 - 8x^2 + 3x^2 + 3x^3 - 7x^2 = x + 2x^3 - 2x - 2x^3 + x$

Now if I put them in order:

$ 3x^3 + x^3 -8x^2 -7x^2 + 3x^2 = 2x^3 - 2x^3 - 2x + x + x$

and combine

$ 4x^3 -12x^2 = 0$

I could add $12x^2$ to both sides and divide by $4x^2$ so it will become $x = 3$

Now I would say there is one solution, also khan academy told me there is either one, infinite or no solution to an equation. But according to Brilliant.org, there are 2 solutions: either 3 or 0. They use the distributive property after combining so that $ 4x^3 -12x^2 = 0$ becomes $ 4x^2(x-3) = 0$. They conclude that either x can be 3 or 0 so one of the fractions become 0 and so will the product, using the zero product property.

It does make perfect sense since I can just put the answer in the original equations and they work... But now I am also confused, isn't this true for most equations and can equations have multiple solutions, maybe even more than 2?

Solution 1:

The statement from Khan academy, that there are either none, one or infinitely many solutions, most likely refers to systems of linear equations.

In this case there are two solutions; in the process of finding your solution $x=3$ you have divided by $4x^2$. But this is only possible if $x\neq0$. So your argument works only if $x\neq0$, meaning that you have to check the case $x=0$ separately.

Solution 2:

The Khan academy article refers to systems of linear equations, meaning that all variables are raised to the power of at most 1. This is a non-linear equation, as it has terms of degree 2 and 3.

Solution 3:

A cubic equation has either $3$ real roots or $1$ real root and $2$ complex roots. Here, we have three real roots, two of which are identical.

\begin{align*} x^3 - 8x^2 + 3x^2 + 3x^3 - 7x^2 &= x + 2x^3 - 2x - 2x^3 + x\\ 4 x^3 - 12 x^2 &= 0\\ (x-0)(x-0)(x-3) & =\frac{0}{4} \implies x\in\big\{0,0,3\big\} \end{align*}