Find all real solutions for the system [MATH]4x^2-40\left\lfloor{x}\right\rfloor+51=0[/MATH] where [MATH]\left\lfloor{x}\right\rfloor[/MATH] represents the floor of $x$.

Find all real solutions for the system [MATH]4x^2-40\left\lfloor{x}\right\rfloor+51=0[/MATH] where [MATH]\left\lfloor{x}\right\rfloor[/MATH] represents the floor of $x$.

My solution:

First, notice that if we rewrite the equality as  [MATH]4x^2+51=40\left\lfloor{x}\right\rfloor[/MATH], we can tell [MATH]\left\lfloor{x}\right\rfloor[/MATH] must be a positive figure.

Second, we let [MATH]x=\left\lfloor{x}\right\rfloor+\{x\}[/MATH], where $\{x\}$ represents the decimal part of $x$.

This transforms the original equation to become:

[MATH]4x^2-40\left\lfloor{x}\right\rfloor+51=0[/MATH]

[MATH]4(\left\lfloor{x}\right\rfloor+\{x\})^2-40\left\lfloor{x}\right\rfloor+51=0[/MATH]

[MATH]4(\left\lfloor{x}\right\rfloor^2+2\left\lfloor{x}\right\rfloor\{x\}+\{x\}^2)-40\left\lfloor{x}\right\rfloor+51=0[/MATH]

[MATH](4\left\lfloor{x}\right\rfloor^2-40\left\lfloor{x}\right\rfloor+51)+8\left\lfloor{x}\right\rfloor\{x\}+2\{x\}^2=0[/MATH]

[MATH](2\left\lfloor{x}\right\rfloor-17)(2\left\lfloor{x}\right\rfloor-3)+2\{x\}(4\left\lfloor{x}\right\rfloor+\{x\})=0[/MATH]

Note that since [MATH]2\{x\}(4\left\lfloor{x}\right\rfloor+\{x\})\ge 0[/MATH], we know [MATH](2\left\lfloor{x}\right\rfloor-17)(2\left\lfloor{x}\right\rfloor-3)+2\{x\}(4\left\lfloor{x}\right\rfloor+\{x\})[/MATH] must be a negative.

Thus, solving [MATH](2\left\lfloor{x}\right\rfloor-17)(2\left\lfloor{x}\right\rfloor-3)+2\{x\}(4\left\lfloor{x}\right\rfloor+\{x\})\le 0[/MATH] for [MATH]\left\lfloor{x}\right\rfloor[/MATH], we see that we have:

[MATH]2\le \left\lfloor{x}\right\rfloor \le 8[/MATH].

Checking for each case of [MATH]\left\lfloor{x}\right\rfloor=2,\,3,\,4,\,5,\,6,\,7,\,8[/MATH], we find only [MATH]\left\lfloor{x}\right\rfloor=2,\,6,\,7,\,8[/MATH] work, which yields

[MATH]x=\frac{\sqrt{29}}{2},\,\frac{3\sqrt{21}}{2},\,\frac{\sqrt{229}}{2},\,\frac{\sqrt{269}}{2}.[/MATH]