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


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.



There is probably a typo here; it should say real solutions.

Assume it has solutions. $$0\leq t<1, t=\frac{1\pm \sqrt{1+3a^2}}{3} \Leftrightarrow 1 \le \sqrt{1+3a^2} < 2 $$ From the fact that $1-\sqrt{1+3a^2}<0$ if $a \neq 0$. From here, we square both sides to get $$0 \le a^2 \le 1$$ This implies $-1 \le a \le 1$.

  • $\begingroup$ Yes i am getting answer with $1\leq\sqrt{1+3a^2}<2$ but i do not understand how you got this one.@MXYMXY $\endgroup$ – Brahmagupta Mar 10 '16 at 9:32
  • 1
    $\begingroup$ @Brahmagupta The smaller value among $\frac{1\pm \sqrt{1+3a^2}}{3}$ would be $\frac{1-\sqrt{1+3a^2}}{3}$, the larger value would be $\frac{1+ \sqrt{1+3a^2}}{3}$ The minimum value would be less or equal to $0$, and the maximum value would be less than $1$ $\endgroup$ – S.C.B. Mar 10 '16 at 9:35

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. $\,-1\le + \sqrt{1+3a^2} < 2\,$ \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}
  2. $\,-1\le - \sqrt{1+3a^2} < 2\,$ \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 $\mathbb R$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.