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 !
Best Answer
By analyzing the first equation, I was able to find all the solutions.
Firstly, we know $3$ things about the floor functions :
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 $[0,\frac{6+2\sqrt{179}}{20}[$.
$$3x^2+x-\frac{7}{2}\ge 1\text{ for }x \ge \frac{-2+2\sqrt{91}}{12}$$
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.
$$3x^2-x-\frac{7}{2}\ge 1\text{ for }x \ge \frac{1+\sqrt{55}}{6}$$
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.
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 $]\frac{1-3\sqrt{19}}{10},0[$.
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...