Let's say I am trying to solve the equation $ax -b\sqrt{x}=c$ such that $a,b,c>0$. Rearranging, squaring and using the quadratic equation yields the solutions $x^*=\frac{2ac+b^2 \pm b\sqrt{b^2+4ac}}{2a^2}$. We know there is only one solution to this equation because the function $ax-b\sqrt{x}$ starts at 0, decreases to a minimum, before monotonically rising. My question is which solution is the real one and which is the extraneous? I suspect the larger root is the real one, but I can't prove that except when $c$ approaches 0.
Determining the extraneous solution to a radical equation
radical-equationsradicalsrootssolution-verification
Best Answer
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 $x$.