radical-equationsrootssolution-verification
$$x=\sqrt{2x+3}$$
If you solved this traditionally you would get $x_1=3$ & $x_2=-1$. But inputting $x=-1$ in $\sqrt{2x+3}$ gives $+1$ or $-1$. The original equation is only valid if $\sqrt{2x+3}=-1$. So does $x=-1$ technically count as a solution or not?
Before you squared the equation, you got it into the form
$$ \frac{ax-c}{b} = \sqrt{x} $$
So, to filter out extraneous roots, all you have to do is verify that $\frac{ax-c}{b} \ge 0$
Or, substituting your solution,
$$ \frac{b^2\pm\sqrt{b^2+4ac}}{2a} \ge 0 $$
Since $a$ and $c$ are both positive, only the version with $+$ is correct.
When solving equations with radicals, don't solve for $x$. Solve for the radicals instead. It makes filtering out extraneous roots trivial because then all you have to do is require all square roots to be non-negative.
In this case, substitute $u=\sqrt{x}$, solve for $u$, retain non-negative values of $u$, then get back to $x$.
Using Maple's Groebner Basis package, we find that $a$ satisfies the equation $$ 256a^8-1024a^7-2080a^6+3520a^5+3105a^4-4060a^3+1158a^2-92a+1=0 $$ Here's the Maple code . . .
Best Answer
If you substitute it back in the original equation and it does not work, it is not a solution.
$$ a = b \Longrightarrow a^2=b^2$$
but
$$a^2 = b^2 \not \!\!\! \implies a = b.$$