# Why is squaring both sides not producing extraneous roots here?

Process $$1$$:

$$2x-1=0$$

$$2x=1$$

$$x=\frac{1}{2}$$

Process $$2$$:

$$2x-1=0$$

$$(2x-1)^2=0$$

$$(2x-1)(2x-1)=0$$

$$2x-1=0$$

$$2x=1$$

$$x=\frac{1}{2}$$

Why aren't we getting extraneous roots in process $$2$$?

• You are getting an extra root: $x=1/2$ is a double root of the quadratic. Sep 24 at 14:55
• @NickD, an extraneous root is not the same as an extra root. An extraneous root is one that does not solve the original equation.
– Paul
Sep 24 at 17:40
• The original equation has one root. If you square it, you have two roots: one of them is extraneous by my definition, but YMMV. Sep 24 at 17:46
• If x= 1 then x^2= 1 which has two roots, 1 and -1. But if x=0 then x^2= which has only x=1 as root. Sep 24 at 22:43

You don't get extraneous roots because you're squaring zero. There is only one value that, when squared, gives zero. If you did it with a non-zero value, you would add extraneous roots, because there would be a second number that has the same square.

• @lonestudent what does that counterexample have to do with what I said?
– Paul
Sep 24 at 17:39
• I have no idea what you’re saying. The problem isn’t that you get the same value when squaring; it’s that there is no other value has a square equal to zero, in contrast to 1, which has 2 values with it has the square.
– Paul
Sep 24 at 18:40
• @lonestudent $0^2 = 0$, but $(-1)^2 = 1$ as well as $(+1)^2=1$ That is the argument. There are 2 numbers that when squared give rise to 1 Sep 24 at 22:09
• @aaa Yes, now it is clear to me the meaning of this sentence: "If you did it with a non-zero value, you would add extraneous roots, because there would be a second number that has the same square" very nice. Sep 25 at 3:32

Because extraneous roots aren't necessarily produced by squaring both sides. They are produced by setting a sequence of logical implications instead of a sequence of logical equivalences. See examples in this answer .

In your case, it turns out that

\begin{aligned} &2x-1=0\\ \Leftrightarrow \quad&2x=1\\ \Leftrightarrow \quad &x=\tfrac{1}{2}\end{aligned}

as well as

\begin{aligned}&2x-1=0\\ \overset{*}\Leftrightarrow \quad&(2x-1)^2=0\\ \Leftrightarrow \quad&(2x-1)(2x-1)=0\\ \overset{**}\Leftrightarrow \quad&2x-1=0\\ \Leftrightarrow \quad&2x=1\\ \Leftrightarrow \quad&x=\tfrac{1}{2}\end{aligned}

In $$*$$ we can use "$$\Leftrightarrow$$" instead of "$$\Rightarrow$$" because $$0$$ has only one square root.

In $$**$$ we can use "$$\Leftrightarrow$$" without the restriction $$x\neq\frac{1}{2}$$ because $$ab=0$$ iff $$a=0$$ or $$b=0$$ (that is, we are not dividing by zero).

The other equivalences are straightforward.

• It's the squaring of both sides that produces a logical implication instead of equivalence. $x=y \rightarrow x^2 = y^2$ isn't reversible. So your first sentence is incorrect.
– Paul
Sep 24 at 17:53
• @Paul Squaring produces a logical implication only if $y\neq 0$. Thus, in general, squaring does not produce extraneous roots. Sep 24 at 19:12
• @Paul I corrected the first sentence. I think it is clearer now. Sep 24 at 19:20

By zero-product property we have that

$$A\cdot B=0 \iff A=0 \quad \lor \quad B=0$$

therefore

$$(2x-1)(2x-1)=0 \iff 2x-1=0 \quad \lor \quad 2x-1=0 \iff 2x-1=0$$

• Short and nice explanation! Sep 24 at 13:50
• @Pedro I've discussed why in this case we don't produce extraneous roots by squaring, which adresses the question posed. I haven't discuss how extraneous roots can be produced i general.
– user
Sep 24 at 13:57
• @lonestudent, user Oh, sorry. It makes sense now. I misunderstood it. I will delete my previous comment. Sep 24 at 14:03