Is this an extraneous solution?

Question

Does $$\sqrt{2x} = \sqrt{5x+3}$$ have an extraneous solution which is $-1$ or is the solution $-1$ ? As I solved, both sides become $\sqrt{-2}$ but negative numbers can't be in the square root.

Answer 1

Generally, we would say this is an extraneous solution. Indeed, if we look at the domains of our functions beforehand, you ought've noticed that

$$\sqrt{2x}\implies x\ge0$$

$$\sqrt{5x+3}\implies x\ge-\frac35$$

Combined, we can only have $x\ge0$, so $x=-1$ is not a solution here.

Answer 2

You can sketch two graphs $y=\sqrt{2x}$ and $y=\sqrt{5x+3}$, and realize that there is no intersection point. So there is no solution.