Will $-b=\sqrt{a}$ always lead to an extraneous solution?

Question

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?

Answer 1

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.