# 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 $0$?

• When you divide each side by $x$, you assume that $x\neq 0$. But, if this is not the case, then ... – Claude Leibovici Dec 25 '15 at 14:32
• The problem is that you can't divide by $x$ if $x=0$. – Matt Samuel Dec 25 '15 at 14:32
• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Dec 25 '15 at 14:37
• In the first way it is asumed $x\ne 0$ which is obvious – Piquito Dec 25 '15 at 14:40

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.

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.

• Which is the wrong step in the 2nd example? – Perceptioner Dec 25 '15 at 15:25
• And does the symbol '⇔' signify something special in 3rd example? – Perceptioner Dec 25 '15 at 15:30
• @Perceptioner There is nothing wrong. Each implication is valid. However, implications are not equivalences. So, the values which satisfy the last line are not the same values which satisfy the first line; in other words, we didn't solve the equation because we didn't establish a sequence of equivalences. So, the question is: what step is not an equivalence? Which respect to the second comment, yes, $\Rightarrow$ means "implies" and $\Leftrightarrow$ means "is equivalent to". See this. – Pedro Dec 25 '15 at 15:47
• @Perceptioner The algebra of your first method is not useless. Instead, it can be used to solve the equation. But you have to be careful. See my Example 4. – Pedro Dec 25 '15 at 16:04
• +1 Always happy for the answers that go that one (or more) step to provide as much explanation (including counter/examples) as possible. Thanks, keep up the good work! – quapka Dec 25 '15 at 23:53

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.