Why does this method to solve a quadratic equation for $x$ omit $x=0$? Here is a simple quadratic equation:
$$9x^2 - 36x = 0$$
We proceed as following:
\begin{align*}
9x^2 & = 36x\\
9x & = 36\\
x & = 4
\end{align*} 
So, we get $x=4$.
But, here's another way:
\begin{align*}
9x^2 - 36x & = 0\\
x(9x-36) & = 0\\
\end{align*}
Therefore, $\boldsymbol{x=0}$ or $x=4$.
What's the problem in the first method that doesn't allow $x$ to be $0$?
 A: In the step going from $9x^2=36x$ to $9x=36$, you divided by $x$.
As you cannot divide by 0, this step is only valid for $x \ne 0$. So you lost the solution $x=0$ with this transformation.
A: To solve an equation, you have to establish a sequence of logical equivalences.
In your first method, you only established a sequence of logical implications. This is the reason why you lost the solution $x=0$.
Remark: In order to solve an equation, an sequence of implications can fails for two reasons:


*

*You lost one (o more than one) "right solution".

*You find one (or more then one) "wrong solution".


*

*Example 1 (provided by you):
$$\begin{align}
&9x^2-36x=0\\
\Rightarrow\quad&9x^2=36x\\
\Rightarrow\quad&9x=36\qquad \text{if } x\neq 0\\
\Rightarrow\quad&x=4\qquad \text{if } x\neq 0\\
\end{align}$$
Here, we lost the right solution $x=0$.

*Example 2:
$$\begin{align}
&x^2+1=0\\
\Rightarrow\quad&(x^2+1)(x^2-1)=0(x^2-1)\\
\Rightarrow\quad&x^4-1=0\\
\Rightarrow\quad&x^4=1\\
\Rightarrow\quad&x=1\\
\end{align}$$
Here, we find the wrong solution $x=1$.

*Example 3 (provided by you too):
$$\begin{align}
&9x^2-36x=0\\
\Leftrightarrow\quad&x(9x-36)=0\\
\Leftrightarrow\quad&x=0\text{ or }9x-36=0\\
\Leftrightarrow\quad&x=0\text{ or }x=4\\\end{align}$$
Here, we lost nothing right and find nothing wrong.

*Example 4:
$$\begin{align}
&9x^2-36x=0\\
\Leftrightarrow\quad&9x^2=36x\\
\Leftrightarrow\quad&9x=36 \text{ if } x\neq 0\qquad \text { or } \qquad x=0\\
\Leftrightarrow\quad&x=4\text { or }x=0\\
\end{align}$$
Here, we solved the equation because we established a sequence of equivalences.
A: You divided by $x$. Note that since you cannot divide by $0$, dividing by $x$ is similar to as asserting that $x$ is not $0$.
Of course, this argument is not rigourous, consider $x^3-x^2=0$. Even when we divide by $x$, we have $x(x-1)=0$, so $0$ and $1$ are solutions to the equation after dividing by $x$.
In summary, the best way is to check if whatever you are dividing by is $0$, if it is, you should try factoring the expression instead.
