<s>[INST] Solve $\lfloor x \rfloor ^2 + 9 \{x\}^2=6x-10, \forall x \in \mathbb{R}$ Solve $$\lfloor x \rfloor ^2 + 9 \{x\}^2=6x-10, \forall x \in \mathbb{R}$$
I tried to:
$$\lfloor x \rfloor ^2 + 9 (x-\lfloor x \rfloor)^2=6x-10$$
$$ 18 x \lfloor x \rfloor -10 \lfloor x \rfloor^2=(3x-1)^2 +9 $$
but I got stuck from here, or I started the exercise wrong? Thank you!
 [/INST] Notice that
$$x\ge \lfloor x\rfloor>x-1$$
$$x^2\ge \lfloor x\rfloor ^2>x^2-2x+1$$
$$x^2+9>9\{x\}^2+\lfloor x\rfloor^2>x^2-2x+1$$
$$x^2+9>6x-10>x^2-2x+1$$
$$4+\sqrt5\approx 6.2>x>4-\sqrt5\approx 1.7$$
Hence $\lfloor x\rfloor\in\{2,3,4,5,6\}$. Now do casework based on each value of $\lfloor x\rfloor$.
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