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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 $2$?

lone student
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tryingtobeastoic
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    You are getting an extra root: $x=1/2$ is a double root of the quadratic. – NickD Sep 24 '21 at 14:55
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    @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 '21 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. – NickD Sep 24 '21 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. – user247327 Sep 24 '21 at 22:43

3 Answers3

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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.

Paul
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    @lonestudent what does that counterexample have to do with what I said? – Paul Sep 24 '21 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 '21 at 18:40
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    @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 – aaaaa says reinstate Monica Sep 24 '21 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. – lone student Sep 25 '21 at 03:32
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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.

Pedro
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  • 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 '21 at 17:53
  • @Paul Squaring produces a logical implication only if $y\neq 0$. Thus, in general, squaring does not produce extraneous roots. – Pedro Sep 24 '21 at 19:12
  • @Paul I corrected the first sentence. I think it is clearer now. – Pedro Sep 24 '21 at 19:20
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By zero-product property we have that

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

therefore

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

user
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    Short and nice explanation! – lone student Sep 24 '21 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 '21 at 13:57
  • @lonestudent, user Oh, sorry. It makes sense now. I misunderstood it. I will delete my previous comment. – Pedro Sep 24 '21 at 14:03