<s>[INST] How to solve $\ x^2-19\lfloor x\rfloor+88=0 $ I have no clue on how to solve this. If you guys have, please show me your solution as well.
$$\ x^2-19\lfloor x\rfloor+88=0 $$
 [/INST] Since $x \geq \lfloor x \rfloor$ for all $x$, you have $x^2 - 19\lfloor x \rfloor + 88 \geq x^2 - 19x + 88$ for all $x$. Hence the graph of $x^2 - 19\lfloor x \rfloor + 88$ lies above that of $x^2 - 19x + 88$. So the function $x^2 - 19\lfloor x \rfloor + 88$ can only be zero where the function $x^2 - 19x + 88 = (x - 11)(x - 8)$ is less than or equal to zero, namely the interval $[8,11]$.
One can then solve $x^2 - 19\lfloor x \rfloor + 88 = 0$ separately on the four ranges in $[8,11]$ where $\lfloor x \rfloor$ is constant: the intervals $[8,9),[9,10],[10,11)$ and the singleton $x = 11$. You get the equations $x^2 - 64 = 0$, $x^2 - 83 = 0$, $x^2 - 102 = 0$, and $x^2 - 121 = 0$ respectively, leading to the four solutions $8, \sqrt{83}, \sqrt{102},$ and $11$.
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