The whole story began when I was developing an easy way to solve complex equations involving mix-up of absolute value, floor, ceiling, rounding, and other functions, as well as polynomials embedded in them.

When I was pretty much done, I started solving different equations like that. I came up with couple equations. One of them is:

$$|x^2 + 3x| + \lfloor2x^2\rfloor + x = 5$$

I ended up with eight numbers. I expected six of them to be extraneous, and two of them to work. However, it turned out, none of the eight numbers worked. So I concluded there are no solutions. Just to be sure, I opened up Wolfram Alpha and wanted to see if there is a solution: http://www.wolframalpha.com/input/?i=abs(x%5E2+%2B+3x)+%2B+floor(2x%5E2)+%2B+x+%3D+5

Wolfram Alpha spitted out $x=2\sqrt{2} - 2$.

I checked the solution that WA gave me, but to my surprise it didn't work! If I plugged it inside wolfram alpha for $x$, it worked, but when I did it on paper, it didn't!

Out of curiosity, I typed in this equation in Wolfram Alpha:

$$\lfloor2x^2\rfloor = x+2$$

And, now I was even more surprised. WA gave me solution $-1$, which IS total NONSENSE. It can be checked here: http://www.wolframalpha.com/input/?i=floor(2x%5E2)+%3D+x%2B2 . Obviously $-1$ doesn't work in that equation.

Most important problem: So, I know that Wolfram Alpha gave me the wrong answer for the second equation, however I am still not sure about the first one. Is the solution WA gave for the first equation correct, or am I right about that there is no solution?


I was indeed "hallucinating". One of the eight numbers I got was actually the solution Wolfram Alpha gave me, although I got it in a different form, and somehow it didn't work when I tested it.

Still, the bug present in the second equation needs to be fixed, I'll report it.

  • 3
    $\begingroup$ Looks like a bug in WA to me. They probably don't get a lot of people trying to solve equations involving floor functions. Note that although $-1$ doesn't work, $-0.999999999999999$ is very close to working; likely the $-1$ is actually rounding. $\endgroup$ – Qiaochu Yuan Jun 3 '17 at 1:44
  • $\begingroup$ This question belongs on the Mathematica.stackexchange site, not this site. $\endgroup$ – David G. Stork Jun 3 '17 at 2:07
  • $\begingroup$ @DavidG.Stork actually not, because I asked "Is the solution WA gave for the first equation correct, or am I right about that there is no solution?" which asked for verification of mathematical solution. $\endgroup$ – KKZiomek Jun 3 '17 at 2:08
  • $\begingroup$ $\texttt{Mathematica}$ has the same problems with those kind of functions. $\endgroup$ – Felix Marin Jun 10 '17 at 2:39
  • $\begingroup$ See my W & A attempt. I wrote $\texttt{Solve[Abs[x^2 + 3 x] + Floor[2 x^2] + x == 5,x]}$ $\endgroup$ – Felix Marin Jun 10 '17 at 2:48

What's happening (for the second one) is that, for example

$$ \lfloor 2(-0.99)^2 \rfloor - (-0.99) = \lfloor 1.9602 \rfloor + 0.99 = 1.99 \approx 2. $$

So what WolframAlpha does, when trying to solve numerically, is get $-0.9999\dots$ (to some chosen precision) which is very close as long as there are only a finite number of '9's. It then rounds (when printing) and gets $-1$ which is no longer a solution.

The first one is a solution:

\begin{align} &\qquad |(2\sqrt 2 - 2)^2 + 3(2\sqrt 2 - 2)| + \lfloor 2(2\sqrt 2 - 2)^2 \rfloor + (2\sqrt 2 - 2) \\ &= |6 - 2\sqrt 2| + \lfloor 1.37 \rfloor + (2\sqrt 2 - 2) \\ &= 5. \end{align}

  • $\begingroup$ I don't think that's what Wolfram Alpha does, although I would agree with you that the floor function gets crazy with discontinuity. $\endgroup$ – KKZiomek Jun 3 '17 at 1:47
  • $\begingroup$ By the way I still need an answer to the last part of my question: "Is the solution WA gave for the first equation correct, or am I right about that there is no solution?" $\endgroup$ – KKZiomek Jun 3 '17 at 1:48
  • 1
    $\begingroup$ Likely it isn't literally computing a limit but arriving at an approximate solution like $-0.999999999999$ and then displaying it as $-1$. Perhaps it defaults to printing floating point numbers to a certain number of decimal places which causes that number to round to $-1$. $\endgroup$ – Erick Wong Jun 3 '17 at 1:52
  • $\begingroup$ Sorry, wasn't trying to claim that WA was taking a limit. $\endgroup$ – Trevor Gunn Jun 3 '17 at 1:54
  • $\begingroup$ @KKZiomek I also agree that $2\sqrt2 - 2$ is a solution. Can you explain what you believe the LHS is for that value of $x$? $\endgroup$ – Erick Wong Jun 3 '17 at 1:55

The function is discontinuous, but if you specify the proper range (e.g., $x > -10$ or $x<20$, to avoid an infinite number of discontinuities), then Mathematica gives the unique solution directly:

Solve[Abs[x^2 + 3 x] + Floor[2 x^2] + x == 5 && x > -10, x]


{{x -> -2 + 2 Sqrt[2]}}


Interestingly, there is no solution for $x<0$:

Solve[Abs[x^2 + 3 x] + Floor[2 x^2] + x == 5 && x < 0, x]




as can be seen by this graph of $|x^2 + 3 x| + \lfloor 2 x^2 \rfloor + x - 5$:

enter image description here

Also, when I enter the equation in free-form WolframAlpha form in Mathematica the unique answer is returned.


Let's solve the first equation, $$f(x) = |x^2+3x| + \lfloor 2x^2 \rfloor + x = 5.$$ We note that when $x \le -3$, the LHS exceeds $5$, since $x^2 + 3x = x(x+3) \ge 0$, and $$\lfloor 2x^2 \rfloor + x - 5 > 2x^2 + x - 6 = (x+2)(2x-3) > 0.$$ When $x \ge 1$, the LHS again exceeds $5$, since in this case $f$ is clearly increasing and we have $f(x) \ge f(1) = 1+3+2+1 = 7 > 5$. So now let us restrict our attention to the interval $x \in (-3, 1)$. On the interval $[0,1/\sqrt{2})$ we have $$f(x) = x^2 + 3x + 0 + x = x^2 + 4x,$$ thus $$f(x) - 5 = (x-1)(x+5)$$ and this admits no roots. On the interval $[1/\sqrt{2},1)$ we have $$f(x) - 5 = x^2 + 4x - 4$$ which admits the unique root in this interval $$\boxed{x = 2(\sqrt{2}-1)}.$$ The negative case is handled similarly, noting that $|x^2 + 3x| = -(x^2 + 3x)$ whenever $-3 < x < 0$, and since $$\lfloor 2x^2 \rfloor = k, \quad x \in \left(-\sqrt{(k+1)/2}, -\sqrt{k/2}\right],$$ we have $$f(x) - 5 = -x^2 - 2x + k-5 = -(x + 1 + \sqrt{k-4})(x + 1 - \sqrt{k-4}).$$ Thus we seek integers $4 \le k \le 8$ such that $-\sqrt{k+1} < -\sqrt{2}(1 \pm \sqrt{k-4}) \le -\sqrt{k}$, or equivalently, $$0 \le k \pm 4\sqrt{k-4} - 6 < 1.$$ Simple trial evaluations yield no such $k$; therefore, the boxed solution is unique.

In Mathematica, the single command

Reduce[Abs[x^2 + 3 x] + Floor[2 x^2] + x == 5, x, Reals]

yields the solution

x == -2 + 2 Sqrt[2].

Typing in

Reduce[Floor[2 x^2] == x + 2, x, Reals]



as expected.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.