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

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

My solution:

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

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

This transforms the original equation to become:

$\displaystyle 4x^2-40\left\lfloor{x}\right\rfloor+51=0$

$\displaystyle 4(\left\lfloor{x}\right\rfloor+\{x\})^2-40\left\lfloor{x}\right\rfloor+51=0$

$\displaystyle 4(\left\lfloor{x}\right\rfloor^2+2\left\lfloor{x}\right\rfloor\{x\}+\{x\}^2)-40\left\lfloor{x}\right\rfloor+51=0$

$\displaystyle (4\left\lfloor{x}\right\rfloor^2-40\left\lfloor{x}\right\rfloor+51)+8\left\lfloor{x}\right\rfloor\{x\}+2\{x\}^2=0$

$\displaystyle (2\left\lfloor{x}\right\rfloor-17)(2\left\lfloor{x}\right\rfloor-3)+2\{x\}(4\left\lfloor{x}\right\rfloor+\{x\})=0$

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

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

$\displaystyle 2\le \left\lfloor{x}\right\rfloor \le 8$.

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

$\displaystyle x=\frac{\sqrt{29}}{2},\,\frac{3\sqrt{21}}{2},\,\frac{\sqrt{229}}{2},\,\frac{\sqrt{269}}{2}.$