By William McCallum

*What are extraneous solutions?*

A while ago I wrote a blog post about solving equations where I talked about seeing the steps in solving equations as logical deductions. Thus the steps

\begin{align*}3x + 2 &= 5\\3x &= 3\\x &= 1\\ \end{align*}

are best thought of as a sequence of if-then statements: If $x$ is a number such that $3x + 2 = 5$, then $3x = 3$; if $3x = 3$, then $x = 1$.

In this post I’d like to explore this way of thinking further to understand the pesky phenomenon of “extraneous solutions.” I put that in quotes because extraneous solutions are not actually solutions at all. Consider the following steps:

\begin{align*}

\sqrt{x} & = -x\\

(\sqrt{x})^2 &= (-x)^2\\

x &= x^2\\

x-x^2 &= 0\\

x(1-x) &= 0 \\

x &= \text{0 or 1}\\

\end{align*}

What’s going on here? It is true that $x = 0$ is a solution of the original equation, because $\sqrt{0} = 0$, but it is not true that $x = 1$ is a solution, because $\sqrt{1} \neq -1$ (we follow the convention here that $\sqrt{}$ denotes the positive square root, so $\sqrt{1} = 1$). Did something go wrong? Then answer is no. The logic is impeccable; every step follows from the previous one, and the concluding statement, $x=0$ or $x=1$, is perfectly true. It so happens that $x \neq 1$, but that doesn’t make the statement “$x = 0$ or $x = 1$” false because in order for an “or” statement to be true we only need one of the things to be true. In short, the statement

If $\sqrt{x} = -x$ then $x = 0$ or $x =1$

is true. However, the converse statement,

If $x = 0$ or $x = 1$ then $\sqrt{x} = -x$,

is not true. In logic, the converse of an if-then statement is the if-then statement with the if part and the then part reversed. The converse of a true statement is not necessarily true. For example, it’s true if you were born in the US then you are a US citizen, but it’s not true that if you are a US citizen then you were born in the US. I myself am a counterexample, since I was born in Australia and then became a naturalized US citizen.

We don’t have to worry about this when we are solving linear equations because all the steps we use to solve linear equations—adding the same number to both sides, multiplying both sides by the same non-zero number—happen to have true converses. It is true $x = y$ then $3x = 3y$ and it is also true that if $3x = 3y$ then $x = y$. This is because the operations of adding a number or multiplying by a non-zero number are reversible; you can just subtract or divide by the same number to get the original equation back again.

However, squaring is not a reversible operation. Although is true that if $x = y$ then $x^2 = y^2$, it is not true that if $x^2 = y^2$ then $x = y$. There are two possibilities if $x^2 = y^2$: either $x = y$ or $x = -y$. Thus the very first step in the solution above had a false converse. It is not necessarily true that if $(\sqrt{x})^2 = (-x)^2$ then $\sqrt{x} = -x$; it could be that $\sqrt{x} = x$. And indeed, this is the possibility that leads to the extraneous solution $x = 1$.

You might be thinking, but hang on, isn’t taking the square root the reverse of squaring? Not quite. Although it is true that $\sqrt{x^2} = x$ when $x$ is positive or zero, it is not true when $x$ is negative. In that case, because $\sqrt{}$ means the positive square root, we have $\sqrt{x^2} = -x$.

The key to understanding most issues that come up when solving equations is to think of the steps in an equation as reasoning steps, not mechanical moves, and to use reasoning language to describe them. Put in the ifs and the thens and think about whether there might be a logical flaw in the reasoning.

**Next Steps**

Can you think of other situations in which “doing the same thing to both sides” leads to misconceptions?

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