Why does this method to solve a quadratic equation for $x$ omit $x=0$?

Question

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 $0$?

Answer 1

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.

Answer 2

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.

Answer 3

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:

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

2. 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.