How to solve those two polynomials which contains a lot of floor function?

Solve for $$x$$ : $$3x^2-2x\lfloor x\rfloor + 4\lfloor x^2\rfloor + x - 4\lfloor x\rfloor-\frac{7}{2}=0$$

Then solve for $$x$$ : $$\lfloor 3x^2-2x\lfloor x\rfloor + 4\lfloor x^2\rfloor + x - 4\lfloor x\rfloor-\frac{7}{2}\rfloor=0$$

And yes, the second equation is the same as the first one but it is inside a floor function.

I don't know how to solve those complicated equation. How would you solve it ?

Thanks for the help !

• First ignore the floors to see where the roots are. Once you have an idea, restrict to intervals between integers, i.e. $x=n+y$, $0\le y<1$. Plotting helps a lot here. – Chrystomath Jul 12 '20 at 11:00
• CAS like Mathematica 12.1.1 can solve both examples. – Mariusz Iwaniuk Jul 12 '20 at 11:25

By analyzing the first equation, I was able to find all the solutions.

Firstly, we know $$3$$ things about the floor functions :

• $$\forall x \in \mathbb R$$, $$x-1\le\lfloor x\rfloor\le x$$
• $$\forall x \in \mathbb R_+$$, $$x^2-x\le x\lfloor x\rfloor\le x^2$$
• $$\forall x \in \mathbb R$$, $$x^2-1\le\lfloor x^2\rfloor\le x^2$$

Ok now, let's use those inequalities to solve the first equation !

Case $$1$$ : $$x\ge 0$$ $$x^2-x\le x\lfloor x\rfloor\le x^2$$ $$-2x^2\le -2x\lfloor x\rfloor\le -2x^2+2x$$ $$3x^2-2x^2\le 3x^2-2x\lfloor x\rfloor\le 3x^2-2x^2+2x$$ $$x^2+4x^2-4\le 3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor\le x^2+2x+4x^2$$ $$5x^2-4+x\le 3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x\le 5x^2+2x+x$$ $$5x^2-4+x-4x\le 3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor\le 5x^2+2x+x-4x+4$$ $$5x^2-3x-4-\frac{7}{2}\le 3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor-\frac{7}{2}\le 5x^2-x+4-\frac{7}{2}$$ $$5x^2-3x-\frac{15}{2}\le 3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor-\frac{7}{2}\le 5x^2-x+\frac{1}{2}$$

Ok now, solve for $$5x^2-3x-\frac{15}{2}=0$$ and for $$5x^2-x+\frac{1}{2}=0$$

We get $$0$$ solution for the second one and $$2$$ for the first one. Caution, the solution must be greater or equal than $$0$$. And the solution which satisfies this rule is $$\frac{6+2\sqrt{159}}{20}$$.

So, if there is a solution greater or equal to $$0$$, $$x$$ needs to be between $$0$$ and $$\frac{6+2\sqrt{159}}{20}$$.

Now we have $$2$$ possibilities : $$\lfloor x \rfloor = 0$$ or $$\lfloor x \rfloor = 1$$

Case $$1.1$$ : $$\lfloor x \rfloor = 0$$

So the function $$3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor-\frac{7}{2}$$ is simplified into $$3x^2+x-\frac{7}{2}=0$$

And we get the solution (greater than $$0$$) : $$\frac{-1+\sqrt{43}}{6}$$ which is equal to $$0$$ when we use the floor function.

Case $$1.2$$ : $$\lfloor x \rfloor = 1$$

After some simplification, we get : $$3x^2-x+4\lfloor x^2 \rfloor-\frac{15}{2}$$

$$\lfloor x^2 \rfloor$$ could be equal to $$1$$ or $$2$$ because here $$1\le x\le\frac{6+2\sqrt{159}}{20}$$ so $$1\le x^2 \le\frac{6+2\sqrt{159}}{20}^2$$

Case $$1.2.1$$ : $$\lfloor x^2 \rfloor=1$$

We get : $$3x^2-x+4-\frac{15}{2}=3x^2-x-\frac{7}{2}$$

There is one solution greater or equal than $$0$$, where the floor is equal to $$1$$ and where the floor of the square is equal to $$1$$. It is : $$\frac{1+\sqrt{43}}{6}$$

Case $$1.2.2$$ : $$\lfloor x^2 \rfloor=2$$

There isn't a solution which satifies every parameters (greater or equal than $$0$$, the floor of the number is equal to $$1$$ and the floor of the square is equal to $$2$$)

Case $$2$$ : $$x\lt 0$$

We get : $$5x^2-x-\frac{15}{2}\le 3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor-\frac{7}{2}\le 5x^2-3x+\frac{1}{2}$$

Ok now, solve for $$5x^2-x-\frac{15}{2}=0$$ and for $$5x^2-3x+\frac{1}{2}=0$$

We get $$0$$ solution for the second one and $$2$$ for the first one. Caution, the solution must be less than $$0$$. And the solution which satisfies this rule is $$\frac{2-2\sqrt{151}}{20}$$.

So, if there is a solution less than $$0$$, $$x$$ needs to be between $$\frac{2-2\sqrt{151}}{20}$$ and $$0$$.

Now we have $$2$$ possibilities : $$\lfloor x \rfloor = -1$$ or $$\lfloor x \rfloor = -2$$

Case $$2.1$$ : $$\lfloor x \rfloor = -1$$

We get : $$3x^2+3x+4\lfloor x^2\rfloor + \frac{1}{2}=0$$

Case $$2.1.1$$ : $$\lfloor x^2\rfloor=1$$

If that's the case, then $$x=-1$$. However, when $$x =-1$$, it's not equal to $$0$$. So it's wrong.

Case $$2.1.2$$ : $$\lfloor x^2\rfloor=0$$

Then we have : $$3x^2+3x + \frac{1}{2}=0$$. And here there is 2 solutions which satisfies everything (less than $$0$$, floor equal to $$-1$$ and floor of the square to $$0$$.

And it's : $$\frac{-3-\sqrt{3}}{6}, \frac{-3+\sqrt{3}}{6}$$

Case $$2.2$$ : $$\lfloor x \rfloor = -2$$

We have : $$3x^2+5x+4\lfloor x^2\rfloor + \frac{9}{2}=0$$

$$\lfloor x^2 \rfloor$$ is equal to $$1$$ because here $$\frac{2-2\sqrt{151}}{20}\le x\le -1$$ so $$1\le x^2 \le\frac{2-2\sqrt{151}}{20}^2$$

We get now : $$3x^2+5x + \frac{17}{2}=0$$. However, there is no solutions.

So finally, there is $$4$$ solutions : $$x=\left\{\frac{-3\pm\sqrt{3}}{6},\frac{\pm 1+\sqrt{43}}{6}\right\}$$

Ok so after a lot of thinking, I found a way to solve for the second equation.

First, we have inside the floor function : $$\lfloor x\rfloor$$ and $$\lfloor x^2\rfloor$$.

This allows us to deduce when we have discontinuities in the function.

For $$x\ge 0$$, we have at $$1$$,$$\sqrt{2}$$,$$\sqrt{3}$$,... discontinuities.

Now, let us recall something I said earlier.

For $$x\ge 0$$ : $$5x^2-3x-\frac{15}{2}\le 3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor-\frac{7}{2}$$

$$5x^2-3x-\frac{15}{2}\ge 1\text{ for }x\ge \frac{6+2\sqrt{179}}{20}$$

So we know for sure that for $$x\ge \frac{6+2\sqrt{179}}{20}$$, $$3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor-\frac{7}{2}\ge 1$$.

The solutions needs to be in the interval $$[0,\frac{6+2\sqrt{179}}{20}[$$.

• For $$x\in[0,1[$$, we have this : $$3x^2+x-\frac{7}{2}$$ (it's easy to show that it is increasing in this interval)

$$3x^2+x-\frac{7}{2}\ge 1\text{ for }x \ge \frac{-2+2\sqrt{91}}{12}$$

But $$\frac{-2+2\sqrt{91}}{12}\gt 1$$, plus because this function is increasing and we know it is equal to $$0$$ at $$x=\frac{-1+\sqrt{43}}{6}$$

The interval $$[\frac{-1+\sqrt{43}}{6},1[$$ is a solution to this equation.

• For $$x\in[1,\sqrt{2}[$$, we have this : $$3x^2-x-\frac{7}{2}$$ (it is increasing in this interval)

$$3x^2-x-\frac{7}{2}\ge 1\text{ for }x \ge \frac{1+\sqrt{55}}{6}$$

But $$\frac{1+\sqrt{55}}{6}\lt \sqrt{2}$$, plus because this function is increasing and we know it is equal to $$0$$ at $$x=\frac{1+\sqrt{43}}{6}$$

The interval $$[\frac{1+\sqrt{43}}{6},\frac{1+\sqrt{55}}{6}[$$ is a solution to this equation.

• For $$x\in[\sqrt{2},\frac{6+2\sqrt{179}}{20}[$$ because $$\frac{6+2\sqrt{179}}{20}\lt\sqrt{3}$$, we have this : $$3x^2-x+\frac{1}{2}$$ (it is increasing in this interval)

If $$x=\sqrt{2}$$ then we would have $$-\sqrt{2}+\frac{13}{2}$$ which is bigger than $$1$$. And because it's increasing, it'll always be bigger than $$1$$. So there is no solutions in this interval.

For $$x\lt 0$$, we have at $$-1$$,$$-\sqrt{2}$$,$$-\sqrt{3}$$,... discontinuities.

$$5x^2-x-\frac{15}{2}\le 3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor-\frac{7}{2}$$

$$5x^2-x-\frac{15}{2}\ge 1\text{ for }x\le \frac{1-3\sqrt{19}}{10}$$

So we know for sure that for $$x\le \frac{1-3\sqrt{19}}{10}$$, $$3x^2-2x\lfloor x\rfloor+4\lfloor x^2\rfloor+x-4\lfloor x \rfloor-\frac{7}{2}\ge 1$$

The solutions needs to be in the interval $$]\frac{1-3\sqrt{19}}{10},0[$$.

• For $$x\in ]-1,0[$$, we have $$3x^2+3x+\frac{1}{2}$$ (it is decreasing from $$-1$$ to $$-\frac{1}{2}$$ and it is increasing from $$-\frac{1}{2}$$ to $$0$$).

This equation is equal to $$0$$ when $$x=\frac{-3\pm\sqrt{3}}{6}$$.

However, we know it's decreasing then increasing. So the intervals $$]-1,\frac{-3-\sqrt{3}}{6}$$ and $$]\frac{-3+\sqrt{3}}{6},0[$$ are others solutions.

• For $$x=-1$$, we have $$\frac{9}{2}\gt 1$$. It's not a solution.

• For $$]\frac{1-3\sqrt{19}}{10},-1[$$ because $$\frac{1-3\sqrt{19}}{10}\gt -\sqrt{2}$$, we have $$3x^2+5x+\frac{17}{2}$$ (it is decreasing in this interval)

It can be shown really easy that $$\forall x\in\mathbb R$$, $$3x^2+5x+\frac{17}{2}\gt 1$$.

So finally, we get : $$x\in\left\{\left]-1,\frac{-3-\sqrt{3}}{6}\right]\cup\left[\frac{-3+\sqrt{3}}{6},0\right[\cup\left[\frac{-1+\sqrt{43}}{6},1\right[\cup\left[\frac{1+\sqrt{43}}{6},\frac{1+\sqrt{55}}{6}\right[\right\}$$

Hope this is the end...

• $3x^2\pm x-\frac 72=3\left(x\pm\frac 16\right)^2-\frac{43}{12}$ so the roots of $3x^2+ x-\frac 72=0$ are $\frac{-1\pm\sqrt{43}}{6}$ and the roots of $3x^2- x-\frac 72=0$ are $\frac{1\pm\sqrt{43}}{6}$. – Alexey Burdin Jul 12 '20 at 13:15
• I don't understand. What do you mean ? – Jotadiolyne Dicci Jul 12 '20 at 13:17
• $x=\frac{\mp 2+\sqrt{43}}{12}$ are not solutions. – mathlove Jul 12 '20 at 13:56
• For the next equation, if $x\ge 0$ and $5x^2-3x-\frac{15}{2}\lt 1$, then $\lfloor x\rfloor=0,1$. If $x\lt 0$ and $5x^2-x-\frac{15}{2}\lt 1$, then $\lfloor x\rfloor=-1,-2$. – mathlove Jul 12 '20 at 14:26
• Nice, but $x=1,\frac{1+\sqrt{55}}{6}$ are not solutions. – mathlove Jul 13 '20 at 4:55