Solve the equation:

$$\left \lfloor 3x-x^2 \right \rfloor = \left \lfloor x^2 + 1/2 \right \rfloor$$

In the solution it writes

We notice that $x^{2}+\frac{1}{2}> 0$ therfore $\left \lfloor x^2 - 1/2 \right \rfloor \geq 0$ . From there $\left \lfloor 3x-x^2 \right \rfloor = n \geq 0$.

So far all of this I understand but than it writes:

But $3x-x^2 \leq \frac{9}{4}$ and $\left \lfloor 3x-x^2 \right \rfloor< 3$.

I dont understand this last part how did they get $3x-x^2 \leq \frac{9}{4}$ ?

  • 4
    $\begingroup$ I fail to see step 1. Say $x=0$ then $x^2+1/2$ has floor zero so $x^2-1/2$ has floor $-1$ which isn't $\ge 0.$ $\endgroup$ – coffeemath Dec 20 '18 at 4:15
  • $\begingroup$ For $3x-x^2\le \frac 94$, note that $-(x-\frac32)^2$ is the underlying quadratic... $\endgroup$ – abiessu Dec 20 '18 at 4:18
  • $\begingroup$ I think the key observation is that $x^2+\frac{1}{2}$ and $3x-x^2$ are a convex and a concave parabola. $\endgroup$ – Mr.Robot Dec 20 '18 at 4:23

If we put $-x^2+3x$ into translated form we get:

$$-(x^2-3x)=-\left[\left(x-\frac 3 2\right)^2-\left(\frac32\right)^2\right]=-\left(x-\frac 3 2\right)^2+\frac94$$

We can now see that this is $x\mapsto x^2$ that has been dilated by a factor of -1 (graphically: reflected about the $x$ axis) and translated $\frac 3 2$ units right and $\frac 9 4$ units up. (In simple terms, the graph is open downwards and the vertex is at $\left(\frac 3 2, \frac 9 4\right)$.) Hence, for all $x$, $-x^2+3x\le \frac 9 4$.


I think there's a typo in the first part of your presented solution (see coffeemath's comment): it should be:

... therefore $\left\lfloor x^2+\frac 1 2\right\rfloor\ge0$. From there...

In any case, I think the solution is directing you to to identify potential cases to investigate: this equation can only hold for floors of 0, 1 and 2.

For example, investigating the floor 0 case:

$$\begin{align} \left\lfloor x^2+\frac 1 2\right\rfloor = 0&\Rightarrow 0\le x^2+\frac 1 2<1\\ x^2+\frac 1 2\ge 0&\Rightarrow x\in\mathbb R\\ x^2+\frac 1 2 < 1&\Rightarrow x\in\left]-\frac{\sqrt 2}{2},\frac{\sqrt2}{2}\right[\\\\ \left\lfloor-x^2+3x\right\rfloor=0&\Rightarrow0\le-x^2+3x<1\\ -x^2+3x\ge0&\Rightarrow x\in\left[0, 3\right]\\ -x^2+3x<1&\Rightarrow x\in\left]-\infty,\frac{3-\sqrt 5}{2}\right[\;\;\bigcup\;\;\left]\frac{3+\sqrt5}{2},\infty\right[ \end{align}$$

Taking the intersection of all these sets gives us part of the solution to the original problem:

$$x\in\left[0, \frac{3-\sqrt 5}{2}\right[$$

Similarly, investigating the floor 1 case gives us that the equation holds for $x\in\left[\frac{\sqrt 2}{2}, 1\right[$, and the floor 2 case gives us that the equation holds for $x\in\left[\frac{\sqrt6}{2},\frac{\sqrt{10}}{2}\right[$.

The solution set to the problem is thus:

$$\left[0, \frac{3-\sqrt 5}{2}\right[\;\;\bigcup\;\;\left[\frac{\sqrt 2}{2}, 1\right[\;\;\bigcup\;\;\left[\frac{\sqrt6}{2},\frac{\sqrt{10}}{2}\right[$$


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