Will $-b=\sqrt{a}$ always lead to an extraneous solution? I'm given the equation $$\sqrt{2k+1}-\sqrt{2k-5}=4$$ and asked to solve for $k$. Wolfram says that this has no solution which is what I concluded as well. But I am wondering if my logic is correct. Somewhere along the way in solving for $k$ we come to the equation: $$\sqrt{2k-5} =-\frac{3}{2}.$$ Immediately I am thinking that this statement can never be true, as $\sqrt{2k-5}$ is the principle (positive) square root of $2k-5$. Hence, it can not yield a negative value. I would love to be able to say that if we ever come to an equation like $$-b = \sqrt{a}$$ for $a$, $b\in \mathbf{R}^+$ we can stop, put our pencils down, and conclude that our original equation has no solution. $$$$ My question is this: can we? If not, what is an equation where we come to something like $-b=\sqrt{a}$, square both sides, ridding the negative, and find a valid solution?
 A: I'll sum up the concepts it seems a few have addressed in the comments.
Setting the square root of something equal to a negative will only result in an imaginary solution, as I'm sure you are aware. Thus, squaring both sides will ALWAYS result in an extraneous solution. So, if you can show that an equation is of the form "square root = a negative," then you know the equation has no real solutions. So yes, your logic is correct.
Good of you to be thinking about equations such as these in a more general sense!
A: If
$$
\sqrt{2k+1}-\sqrt{2k-5}=4
$$
the
$$
6=(2k+1)-(2k-5)=(\sqrt{2k+1}-\sqrt{2k-5})(\sqrt{2k+1}+\sqrt{2k-5})
=4(\sqrt{2k+1}+\sqrt{2k-5}).
$$
Hence we obtain the system
$$
\sqrt{2k+1}-\sqrt{2k-5}=4 \\
\sqrt{2k+1}+\sqrt{2k-5} = \frac{3}2
$$
This implies that $\sqrt{2k-5}=-\frac{5}{4}$
IMPOSSIBLE!
Hence, no solution.
A: 
$$\sqrt{2k+1}-\sqrt{2k-5}=4$$ Somewhere along the way in solving for $k$ we come to the equation $$\sqrt{2k-5} =-\frac{3}{2}.$$


I would love to be able to say that if we ever come to an equation like $$\sqrt{a}=-b$$ for $a$, $b\in \mathbf{R}^+$ we can stop, and conclude that our original equation has no solution.

To be clear, I think you are asking whether $$\sqrt{f(x)}=-b\quad\text{and}\quad b>0\tag1$$ can have a solution.
The answer is No, simply by definition: since $\sqrt{f(x)}$ is defined to never be negative, no complex (including real) value of $x$ can ever satisfy $(1),$ that is, make it true. As such, that equation in $(1)$ is called an inconsistent equation.
