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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?

Chris Christopherson
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  • Yes, we can.${}{}{}$ – markvs Dec 14 '21 at 22:48
  • @markvs haha okay great. But why? – Chris Christopherson Dec 14 '21 at 22:57
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    You already said it, $\sqrt{t}$ means, by definition, the unique positive square root of $t$ whenever $t > 0$ (it means zero when $t = 0$) and it means $\sqrt{-t} i$ when $t < 0$ and $i$ is the complex unit. In part, we don't allow for other definitions because $\sqrt{z}$ for $z$ complex cannot be defined well everywhere, and, in fact, depending how you cut the (complex) plane, you get different definitions.... – William M. Dec 14 '21 at 23:04
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    ...There is a theorem of the sort that states if $D$ is a region in the complex plane that contains the positive (real) axis and it has no holes, then $\sqrt{z}$ can be extended uniquely on $D$ such that $\sqrt{z}$ coincides to the usual (positive) square-root when $z$ is in the positive (real) axis. (The largest of a typical region $D$ will then be something like all the complex plane except the negative (real) axis.) – William M. Dec 14 '21 at 23:04
  • Assuming you are restricted to working in the reals, then yes you can stop and conclude that no solutions exist. Formally, you can say "Assume that a solution exists, then with this working, we get to $0 > -b = \sqrt{ a } > 0$ which is a contradiction. Hence no solutions exist". $ \quad$ (Answer is different if you're working in the complex) – Calvin Lin Dec 14 '21 at 23:54
  • If we're working in real numbers, then $\sqrt{a}$ where $a\ge 0$ is defined as the unique non-negative number $b$ such that $b^2=a$. So it simply follows from the way things are defined that square roots are non-negative. – bjorn93 Dec 15 '21 at 00:22

3 Answers3

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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!

Clyde Kertzer
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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.

Yiorgos S. Smyrlis
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  • Ok. But my question pretty much comes down to why it is impossible and why we do not need to proceed further. As you can see I have already attempted to anwer these questions: is my logic correct? – Chris Christopherson Dec 14 '21 at 22:56
  • This answers your question, "square both sides, ridding the negative, and find a valid solution?" - You can't get a "valid solution" when the right hand side of a square root is negative. Impossible. – NoChance Feb 23 '23 at 14:20
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$$\sqrt{2k+1}-\sqrt{2k-5}=4$$ 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.

ryang
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