Alright, so I have $\sqrt{x-15} = 3-\sqrt{x}$. I first square both sides to get $x-15 = (3-\sqrt{x})(3-\sqrt{x})$ which simplifies to $x-15 = 9 -6\sqrt{x} + x$.

I solved for $x$ and got $x = 16$, however, when I plug it in, the equation doesn't work. Why does this happen?

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    $\begingroup$ Because $x^2=y^2$ does not imply the same thing as $x=y$. $\endgroup$
    – JMoravitz
    Commented Jun 16, 2016 at 1:31
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    $\begingroup$ Notice that when $x = 16$, you get $\sqrt{16-15} = 1$ yet $3 - \sqrt{16} = 3-4=-1$...but you squared the original equation then solved. Indeed it is true that $1^2 = (-1)^2 = +1$! $\endgroup$
    – Jared
    Commented Jun 16, 2016 at 1:33
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    $\begingroup$ As an aside, one can see that there are no solutions by rearranging. $\sqrt{x-15}+\sqrt{x}=3$. In order to have all terms remain real, $x\geq 15$. However in doing so, $\sqrt{x-15}+\sqrt{x}\geq \sqrt{x}\geq \sqrt{15}>\sqrt{9}=3$ so we see that $\sqrt{x-15}+\sqrt{x}\neq 3$ for any real value of $x$. $\endgroup$
    – JMoravitz
    Commented Jun 16, 2016 at 1:39
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    $\begingroup$ Related: Why can't you square both sides of an equation?, Is there a name for this strange solution to a quadratic equation involving a square root? (and probably you can find other questions on the same topic). $\endgroup$ Commented Jun 16, 2016 at 4:21
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    $\begingroup$ By the way, the solution you came up with is in fact a true solution for $-\sqrt{x-15}=3-\sqrt x$. Generally, any spurious solutions caused by squaring will be true solutions for a variation on the original equation with a different choice of signs for the square root. $\endgroup$ Commented Jun 16, 2016 at 5:35

9 Answers 9


You did good work in doing what many students forget to do - checking that your proposed solution actually is a solution.

Whenever you square both sides of an equation, you potentially introduce new solutions that may or may not be solutions of the original equation. Consider the simple example $x=1$. Obviously $x=1$ is the solution, but squaring both sides gives $x^2 = 1$, which has solutions $x=\pm 1$. Over $\mathbb{R}$ it is most definitely not true that $-1=1$, so $x=-1$ cannot be a solution here.

You should always be careful and check your proposed solutions by plugging them back into the original equation like you have done here.

Indeed, your equation actually has no solutions, which is what you should conclude when asked to solve it.

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    $\begingroup$ I think the term is extraneous solution $\endgroup$
    – Frank
    Commented Jun 16, 2016 at 1:36
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    $\begingroup$ @Frank Yep, extraneous/spurious solutions are used to describe these non-solutions. $\endgroup$ Commented Jun 16, 2016 at 1:37
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    $\begingroup$ If anyone is curious how W|A solves it: imgur.com/tTNC1X1 @Ovi $\endgroup$ Commented Jun 19, 2016 at 0:53

Before doing some computation, you should check the condition of variable. Here, you have the conditions $\begin{cases} x - 15 &\geq& 0, \\ 3 - \sqrt{x} &\geq& 0, \\ x \geq 0. \end{cases}$

You will get that $x \geq 15$ and $0 \leq x \leq 9$. So, you can conclude that the equation has no solution.

  • $\begingroup$ Where is X >= 0 stated? $\endgroup$
    – Tony Ennis
    Commented Jun 18, 2016 at 12:33
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    $\begingroup$ @TonyEnnis The presence of $\sqrt x$ implies that $x \ge 0$. $\endgroup$
    – Théophile
    Commented Jun 18, 2016 at 16:25
  • $\begingroup$ @Théophile so no i? $\endgroup$
    – Tony Ennis
    Commented Jun 18, 2016 at 23:02
  • $\begingroup$ @TonyEnnis We are implicitly working over the real numbers. $\endgroup$
    – Théophile
    Commented Jun 19, 2016 at 4:20

What you have done with your work is prove:

(1) If $x$ satisfies that equation, then $x=16$.

This is not the same thing as

(2) $x$ satisfies that equation if and only if $x = 16$.

Typically, "solving" an equation means to produce a statement like (2).

However, generally, the method of starting with the thing you're trying to solve and making algebraic simplifications only winds up proving something like (1), and additional work is needed to determine the correct statement like (2).

A typical form of that additional work is to check each possibility of the conclusion (here, $x=16$) to see which are actually solutions. Here, we eliminate the only possibility and get

$x$ satisfies that equation if and only if (contradiction).

or, more naturally phrased,

That equation has no solutions for $x$.

One case where additional work is not needed is if every step is "reversible"; i.e. your work is a chain of if and only if steps: e.g. subtraction is reversible, and

$3x + 1 = 5x - 2$ if and only if $1 = 2x - 2$

A lot of basic algebra exercises happen to be solved through reversible steps, which is why you can get the right answer without actually learning how to reason correctly.

Squaring is (generally) not reversible.


Did you check the domain first?

For $\sqrt{x-15}$ to have a real value, $x-15$ must be non-negative, so $x\ge 15$.

Then the LHS is non-negative, so the RHS needs to be non-negatve, too.

And for $3-\sqrt{x}$ to be greater than (or equal to) $0$, $\sqrt x$ must not exceed $3$, hence $x\le 9$.

Conjunction of those requirements $$15 \le x \le 9$$ implies the equation has NO real solution.


Compare Wolfram Alpha's plots of LHS vs. RHS of your original equation


and squared:



Suppose you were given an equation $$\sqrt x = -1\qquad(*)$$ You squared both sides to get $$x = 1$$ and found out the resulting $x$ does not satisfy (*).

Would you be surprised then?
If not, why are you surprised now? :-)


A) In Mathematics if you start with a true statements, apply various rules, you will always get other true statements. However if you start from a false statement, then the end result might be true or false.

For example suppose we have $x^2 = 2x$. This is true when $x = 2$. We can square both sides to get $x^4 = 4x^2$, which is also true when $x = 2$.

Suppose we have $x^2 = 2x$ again. This is false when $x = 1$. We can square both sides to get $x^4 = 4x^2$ which might be true or false when $x = 1$. In this case the $x^4 = 4x^2$ is false when $x = 1$.

Suppose we have $x^2 = 2x$ again. This is false when $x = -2$. We can square both sides get $x^4 = 4x^2$ which might be true or false when $x = -2$. In this case the $x^4 = 4x^2$ is true when $x = -2$.

Thus giving examples of the first two sentences.

B) Here you started with $\sqrt{x-15} = 3 - \sqrt{x}$. This is false when $x = 16$, however when you apply various rules you will get $x = 16$, which is true when $x = 16$. Here you started with something that is false, and derived something that is true! But there is no contradiction because from a false we can get a true.

You might ask what is the point of manipulating an expression if the end result could be true or false regardless whether the original was true or false.

C) Well, if you start with a true statements, apply various rules, you will never get a false statement. So if you start with a statement that might be true or false, apply various rules, and the end result is false, then the original had to be false as well.

For example if we start with $\sqrt{x-15} = 3 - \sqrt{x}$ and suppose we do not know if it is true or false when $x = 5$. I.e. we do not know whether $\sqrt{5-15} = 3 - \sqrt{5}$ is true. We could square both sides and get $5 = 16$. The final result was false, thus $\sqrt{5-15} = 3 - \sqrt{5}$ was false as well, i.e. $\sqrt{x-15} = 3 - \sqrt{x}$ is false when $x = 5$.

Now you could do this for any $x$ except $16$, showing none of these values work. So when you got $x = 16$, you were not showing that $x$ is $16$, you were showing that no other value could be a solution!

When you substituted $x = 16$ and got $1 = -1$, you were showing $x = 16$ does not work either.

Thus manipulating an equation is useful for eliminating values. Once you eliminated some values, you should substitute all values left over back into the original see whether they work. (Unless you used reversible steps, then I can show that substitution is not needed).

  • $\begingroup$ What do you mean by "the latter is false" on the third paragraph? What sentence is "the latter"? $\endgroup$
    – Pedro A
    Commented Jun 17, 2016 at 23:31
  • $\begingroup$ I meant $x^4 = 4x^2$ is false when $x = 1$. $\endgroup$ Commented Jun 18, 2016 at 1:11
  • $\begingroup$ Indeed, I must have misread something. But now I noticed something else: can you reread your second paragraph? It seems to claim $1 = 4$, you probably had a typo or something. $\endgroup$
    – Pedro A
    Commented Jun 20, 2016 at 10:46
  • $\begingroup$ @Hamsteriffic Thanks, I have updated it. (Also you did not misread it, I just updated it the last time as well). $\endgroup$ Commented Jun 20, 2016 at 22:16
  • $\begingroup$ Nice :D excellent answer, by the way. +1 $\endgroup$
    – Pedro A
    Commented Jun 20, 2016 at 23:02

Part of this phenomenon is helped by looking at a graph -

enter image description here

Note - I've used this for more complex trig equations, in which multiple false solutions must be discarded. A graph helps show where results of say -1 and 1 are both squared and of course, become equal. For this question, it's the same thing, only above you can see that the 2 sides of this equation are +/- at a single pair of points.

To the down-voter - Math is tough for many students who have trouble with certain concepts that are best illustrated visually. I've worked with students trying to explain the nature of extraneous trigonometry solutions, and found that while all the other explanations here are 100% correct, that a graph of each side of the equation gave these students a way of seeing how the squaring process injected these incorrect solutions. It was the case where an X value in one side produced +1 but the other side, -1. It was with this process that these students were satisfied. This may just be 1 of 10 students in a class, but my goal is to reach as many as I can.


Let's keep this simple. Suppose you start with $x = -3$. If you square both sides, you get

\begin{align} x^2 &= 9 \\ x^2 - 9 &= 0 \\ (x-3)(x+3) &= 0 \\ x &\in \{3, -3\} \end{align}

So we acquired the extra solution, $x = 3$, which doesn't solve the original equation.

Another way to look at this problem is to note that $3 = -3$ is FALSE, but when you square both sides you get $9 = 9$, which is TRUE.

The point is that squaring both sides of an equation sometimes creates a new equation with more solutions than the original equation. Extra solutions are usually called extraneous roots.

In you equation,$\sqrt{x-15} = 3-\sqrt{x}$, you got $x = 16$. But when you substituted $16$ for $x$ and simplified, you got $1 = -1$. Of course this is FALSE, but the equation you get when you square both sides, is TRUE.


$x = 16$ is an extraneous solution. That is, a solution to a related equation but is not a valid solution to the equation in question.


$$\sqrt x = -1$$

You would naturally square both sides and find the solution $x = 1$. But this does not apply, as we can clearly see by verifying that $\sqrt 1 \ne -1$. In fact, there's no solution in the reals. The extraneous solution arises as a result of squaring each side. The resulting equation could result from squaring each side of any of these:

$$\sqrt x = 1$$ $$\sqrt x = -1$$ $$-\sqrt x = 1$$ $$-\sqrt x = -1$$

And so any solution that you receive from solving $x = 1$ could actually be a solution to any of those, or none of them.

Extraneous roots don't arise only from root equations, though. Consider also:

$$\frac{x^2}x = 0$$

You might choose to multiply both sides by x, and then solve $x^2 = 0$, but the solution $x = 0$ is extraneous since it is excluded from the domain of the equation. Again, there is no soultion in the reals.

But sometimes there are. Consider:

$$\frac{x^2 + x - 6}{x - 2} = 0$$ $$x^2 + x - 6 = 0$$ $$(x + 3)(x - 2) = 0$$ $$x = -3, x = 2$$

But this time, $x = 2$ is extraneous because it is excluded from the domain, but $x = -3$ is a valid root.

So, long story short, going back and checking your work is important not just because you're fallible, but also because it's the easiest way to identify a phony solution that math decided to troll you with.


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