I ran across this equation...

$\sqrt {2x+6}+4=x+3$

Without thinking, I solved for x in the following way:

$\sqrt {2x+6}+4=x+3$

Subtract 4 from both sides.

$\sqrt {2x+6}=x-1$

Square each side.


"FOIL" and rearrange.

$ x^2-4x-5=0 $



If you plot the functions you can very clearly see there is only one solution, x=5, which of course stems from the fact that the radical is does not have the plus/minus sign so only half of the "parabola" is considered. But my problem with this is that it should be possible to show algebraically that there is only one solution (other than plugging the numbers in and checking). My thinking is that there is some logical mistake in the way I solved for x, but I can't find it. Moreover, there are of course equations that do intersect twice ($\sqrt {4x+6}+1=x+3$ at $x=\sqrt 2 , -\sqrt 2$ for example), and the exact same steps work in that case.

Any thoughts as to what the differences are between the two cases would be appreciated.

  • $\begingroup$ In the original equation, you have the implicit condition that $2x+6\geqslant 0$ and $x-1\geqslant 0$. $\endgroup$ – user587192 Nov 14 '18 at 3:34
  • $\begingroup$ Ben, actually when you square an equation you end up with the possibility of getting extraneous roots which you need to eliminate because these extraneous roots don't actually satisfy the original equation. $\endgroup$ – Akash Roy Nov 14 '18 at 3:42

"$\sqrt{2x +6}=x-1$" is true if and only if "2x+6=(x-1)^2 AND $x\geq 1$".

Basically when we do math there are "if and only if" steps where each statement is logically equivalent to the next statement, and there "implies" steps where one statement implies the next statement but not the converse. When you square both sides of an equality, then you are doing an "implies" step.

If you always use "if and only if" steps, then you will be able to conclude that $x=5$ without checking cases.

Here is how we do the proof using only "if and only if" steps.

Proof: The following are equivalent:

  1. $\sqrt{2x +6} +4 = x+3$
  2. $\sqrt{2x +6} = x-1$
  3. $2x+6 = (x-1)^2$ and $x\geq1$
  4. $2x+6 = x^2-2x+1$ and $x\geq1$
  5. $0=x^2-4x-5$ and $x\geq1$
  6. $0=(x-5)(x+1)$ and $x\geq1$
  7. ($x=5$ or $x=-1$) and $x\geq1$
  8. x=5.

I’m not sure if this is what you mean by see it algebraically, but $\sqrt{2x+6}=x-1\geq 0$ implies $x\geq 1$. This would automatically eliminate $x=-1$.


The issue comes in with squaring both sides of the equation. This introduces a second, extraneous solution to the equations.

This is why it's important to check both the solutions you get in cases like these - one will work, and the other won't.

Your way of solving for $x$ is indeed valid, it's just a method that sometimes makes extraneous solutions like these. Checking your work is important for that reason.

  • $\begingroup$ I've been curious about these "extraneous" solutions for awhile now. For instance, x=1 becomes x^2=1, and then x can be -1 or 1. I hadn't realized you end up needing to put more constraints on problems when doing things like that. $\endgroup$ – Ben Nov 14 '18 at 3:53

At the very beginning, you are working with the square root function, which (as a real-valued function of a real variable) is undefined when its argument is negative. In other words, any solution to the equation $$ \sqrt{2x+6}+4 = x+3 $$ must satisfy $$ 2x + 6 \ge 0, $$ as the radical expression will not be defined otherwise. But $$ 2x+6 \ge 0 \iff x \ge -3. $$ So, as you work through the problem, you automatically know that any solution you obtain must be greater than $-3$.

Next, in step where you isolate the radical expression, note that $\sqrt{2x+6}$ is, by definition, a nonnegative number. Hence $$ 0 \le \sqrt{2x+6} = x-1. $$ In other words, your solution must satisfy the relation $$ x-1 \ge 0 \iff x > 1. $$ So now you know that any solution must be greater than 1 (which makes the condition $x\ge -3$ redundant). The rest of the computation is fine, but when you get to the end, you have two possible solutions ($x=-1$ and $x=5$), only one of which satisfies the inequalities established earlier.

The short version is as follows: when you are working with functions which are not defined for all real numbers, you have to be careful to track which solutions actually give you expressions in the domains of those functions. You can either do this carefully as you go, or check for extraneous solutions at the end by "plugging them in."


By reading $\sqrt{2x+6~}$ as the non-negative square root, then the squaring step weakens the constraints on $x$ because it is not bijective.

  • $\sqrt {2x+6~}=x-1$ entails $2x+6=(x-1)^2$, however
  • $2x+6=(x-1)^2$ entails $\sqrt {2x+6~}=x-1$ or $\sqrt{2x+6~}=1-x$.
    • Thus the solutions to the squared formula are $5$ or $-1$.

For the solution to the original statement, use that

  • $\sqrt {2x+6}=x-1$ if and only if both $2x+6=(x-1)^2$ and that $x-1\geq 0$ .
    • Thus the only solution is $5$.

I thank you all for your responses. I really rather wish that I could make two of your answers the "accepted" ones.

Eevee Trainer used the phrase "extraneous solutions," which led me to Wikipedia. All of you have provided good insight into the question, but for the interest of people who may read this thread, I will sum up everything in a little more general way.

Not all operations are "invertible." Wikipedia gives the following example:

$x-2=0$, which is true for $x=2.$

However, multiplying through by zero gives $0=0$, which is true for all x values.

This issue occurs because you cannot undo multiplying by zero. Put more precisely, $x-2=0 \implies 0=0$, which is distinctly NOT biconditional. In order to make it biconditional, more restrictions need to be applied in the conclusion.

There are a number of ways this sort of thing can happen, but it's commonly due to mupltipiers being equal to zero for some x value.


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