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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 !

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    $\begingroup$ 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. $\endgroup$ – Chrystomath Jul 12 '20 at 11:00
  • $\begingroup$ CAS like Mathematica 12.1.1 can solve both examples. $\endgroup$ – Mariusz Iwaniuk Jul 12 '20 at 11:25
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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...

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    $\begingroup$ $3x^2\pm x-\frac 72=3\left(x\pm\frac 16\right)^2-\frac{43}{12}$ 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 $\frac{1\pm\sqrt{43}}{6}$. $\endgroup$ – Alexey Burdin Jul 12 '20 at 13:15
  • $\begingroup$ I don't understand. What do you mean ? $\endgroup$ – Jotadiolyne Dicci Jul 12 '20 at 13:17
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    $\begingroup$ $x=\frac{\mp 2+\sqrt{43}}{12}$ are not solutions. $\endgroup$ – mathlove Jul 12 '20 at 13:56
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    $\begingroup$ 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 $\lfloor x\rfloor=-1,-2$. $\endgroup$ – mathlove Jul 12 '20 at 14:26
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    $\begingroup$ Nice, but $x=1,\frac{1+\sqrt{55}}{6}$ are not solutions. $\endgroup$ – mathlove Jul 13 '20 at 4:55

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