= Extraneous solution
An <extraneous solution> satisfies a transformed <equation> but fails the original <equation> or its restrictions. The term describes a candidate generated during the calculation, rather than a true solution that can be kept in the final <solution set>.
Consider the real <equation>
$$
\sqrt{x}=x-2.
$$
The <square root> requires $x\geq0$. Because the <square root> is nonnegative, any solution must also have $x-2\geq0$, so $x\geq2$. Squaring gives
$$
x=(x-2)^2
\quad\Longleftrightarrow\quad
x^2-5x+4=0
\quad\Longleftrightarrow\quad
(x-1)(x-4)=0.
$$
The squared <equation> has candidates $x=1$ and $x=4$. At $x=4$, the original equality is $2=2$. At $x=1$, it is $1=-1$, which fails: this candidate is an <extraneous solution>. Thus the original <solution set> is exactly $\{4\}$.
\Image[/squaring-equation-extraneous-candidate.png]
{title=Squaring removes a sign distinction and creates a second intersection at $x=1$, which fails the original <equation>}
On the restricted <domain of a function> $x\geq2$, both original sides are nonnegative, so squaring is reversible there and the two <equations> are equivalent. Retaining the sign restriction or checking every candidate in the original <equation> achieves the same final <solution set>. The general need to check candidates after raising both sides to a power is explained in https://openstax.org/books/college-algebra-2e/pages/2-6-other-types-of-equations[College Algebra 2e, §2.6: Solving Radical Equations].
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