extra solutions to radical equations can someone explain why when a radical equation is solved some solutions don't work? Is there a rule when a certain transformation is performed on the equation it gives extra solutions? thanks very much
 A: The steps you may be used to for solving an equation are one-to-one. If all I know about $x$ is that $2x+1=0$, then
$$\begin{align}
&&2x+1&=0\\
&\iff&2x&=-1\\
&\iff&x&=-\frac12
\end{align}$$
Each step is an implication but so too would be the steps in the reverse order.
If all I know about $x$ is that $\sqrt{2-x}=x$ then I might start by squaring both sides:
$$\begin{align}
&&\sqrt{2-x}&=x\\
&\implies&2-x&=x^2\\
\end{align}$$
but already this cannot be reversed. To take the square root of the second line would leave us with $\sqrt{2-x}=|x|$, which is not what we started with. So the one-sided implication symbol can be used to emphasize that there is no going back. Continuing,
$$\begin{align}
&&\sqrt{2-x}&=x\\
&\implies&2-x&=x^2\\
&\iff&0&=(x+2)(x-1)
\end{align}$$
So we've found that if any $x$ at all satisfies $\sqrt{2-x}=x$, then it would have to be $-2$ or $1$. But of course, that has merely narrowed it down from infinitely many possible $x$-values down to two possible $x$-values. It's up to you do a little something more to actually solve the original equation.
A: In general, when a transformation is one-to-one (or injective), then it is reversible, so results in an equivalent equation. The problem with some radical equations is (for example) that $(-1)^2=1=1^2.$ That is, squaring both sides of a false equation (like $-1=1$) may result in a true equation, so we get extraneous solutions to deal with.
