<s>[INST] If $a\in R$ and the equation $-3(x-\lfloor x \rfloor)^2+2(x-\lfloor x \rfloor)+a^2=0$ has no integral solution,then all possible values of $a$ If $a\in R$ and the equation $-3(x-\lfloor x \rfloor)^2+2(x-\lfloor x \rfloor)+a^2=0$ has no integral solution, then all possible values of $a$ lie in the interval 
$(A)(-1,0)\cup(0,1)$
$(B)(1,2)$ 
$(C)(-2,-1)$
$(D)(-\infty,-2)\cup(2,\infty)$

My try:
Let $x-\lfloor x \rfloor  = \{x\}= t$, where $ 0 \leq \{x\}<1\Rightarrow 0\leq t<1$. Then
$$\Rightarrow -3t^2+2t+a^2 = 0\Rightarrow 3t^2-2t-a^2 = 0$$
$$\displaystyle \Rightarrow t = \frac{2\pm \sqrt{4+12a^2}}{6} = \frac{1\pm \sqrt{1+3a^2}}{3}$$
I do not know how to solve further.
 [/INST] First, let us see for which values of $a$ there will be solutions:

Since $\,0\leq t = \left\lfloor x \right\rfloor<1\,$ we have
\begin{align}
t &= \dfrac{1\pm \sqrt{1+3a^2}}{3}  \in \left[0,1\right) 
&\implies&& 0\le1\pm\sqrt{1+3a^2} < 3
\\
&\big(\text{subtract } 1 \text{ from each inequality}\big)
&\implies&& -1\le\pm\sqrt{1+3a^2} < 2
\end{align}
Now consider two cases separately:


*

*$\,-1\le + \sqrt{1+3a^2} < 2\,$
\begin{align}
&-1\le \sqrt{1+3a^2} < 2&\implies&& 0\le\sqrt{1+3a^2} < 2
\\
&\big(\text{square both sides of each inequality}\big)
&\implies&& 0\leq 1+3a^2 <4
\\
&\big(\text{subtract } 1\big)
&\implies&& 0\leq 3a^2<3
\\
&\big(\text{divide by } 3\big)
&\implies&& 0\leq a^2 < 1
\\
&&\implies&& \bbox[1ex, border:solid 1.5pt #e10000]{a\in\left(-1,1\right)}
\end{align}

*$\,-1\le - \sqrt{1+3a^2} < 2\,$
\begin{align}
&-1\le- \sqrt{1+3a^2} < 2&\implies&&  0 < \sqrt{1+3a^2} \leq 1
\\
&\big(\text{square both sides of each inequality}\big)
&\implies&& 0\lt 1+3a^2 \leq 1
\\
&\big(\text{subtract } 1 \text{ again}\big)
&\implies&& -1\lt 3a^2 \leq 0
\\
&&\implies&& \bbox[1ex, border:solid 1.5pt #e10000]{a = 0}
\end{align}



Second, find which values of $a$ correspond to the integral solution of the equation. The answer will be complement to the set of these values in $\mathbb R$.
</s>