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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 $2$?

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    $\begingroup$ You are getting an extra root: $x=1/2$ is a double root of the quadratic. $\endgroup$
    – NickD
    Sep 24 at 14:55
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    $\begingroup$ @NickD, an extraneous root is not the same as an extra root. An extraneous root is one that does not solve the original equation. $\endgroup$
    – Paul
    Sep 24 at 17:40
  • $\begingroup$ The original equation has one root. If you square it, you have two roots: one of them is extraneous by my definition, but YMMV. $\endgroup$
    – NickD
    Sep 24 at 17:46
  • $\begingroup$ 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. $\endgroup$
    – user247327
    Sep 24 at 22:43
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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.

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    $\begingroup$ @lonestudent what does that counterexample have to do with what I said? $\endgroup$
    – Paul
    Sep 24 at 17:39
  • $\begingroup$ 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. $\endgroup$
    – Paul
    Sep 24 at 18:40
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    $\begingroup$ @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 $\endgroup$ Sep 24 at 22:09
  • $\begingroup$ @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. $\endgroup$ Sep 25 at 3:32
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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.

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  • $\begingroup$ 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. $\endgroup$
    – Paul
    Sep 24 at 17:53
  • $\begingroup$ @Paul Squaring produces a logical implication only if $y\neq 0$. Thus, in general, squaring does not produce extraneous roots. $\endgroup$
    – Pedro
    Sep 24 at 19:12
  • $\begingroup$ @Paul I corrected the first sentence. I think it is clearer now. $\endgroup$
    – Pedro
    Sep 24 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

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

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    $\begingroup$ Short and nice explanation! $\endgroup$ Sep 24 at 13:50
  • $\begingroup$ @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. $\endgroup$
    – user
    Sep 24 at 13:57
  • $\begingroup$ @lonestudent, user Oh, sorry. It makes sense now. I misunderstood it. I will delete my previous comment. $\endgroup$
    – Pedro
    Sep 24 at 14:03

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