Quadratic equation involving floor function. If equations $x^2-3x+4=0$ and $ 4x^2-2\lfloor3a+b\rfloor x+b=0\space (a,b\space\epsilon\space R) $ have a common root then the complete set of values of $a$ is ?
I have not yet been able to develop an approach to this problem. How do I go about it?
Edit :
I was able to solve and obtain the common root as $ x= \frac{b-16}{2\lfloor3a+b\rfloor -12} $.
After this, how do I obtain the range of $a$?
 A: The roots to the first quadratic are $x_{\pm}=\frac{3}{2}\pm \frac{i\sqrt{7}}{2}$; in terms of the second quadratic; my advice would be to solve
$$
4x^{2}-2(3a+b)x+b=0
$$
ignroing the floor; once you have a relationship between the roots to the first and the secodn implement the propertiers of the floor to come up with possible values of $a$ and $b$.
A: Begin first by noting that the roots of $x^2-3x+4$ are complex conjugates, $\alpha_1$ and $\alpha_2$, and that the coefficients in $4x^2-2\lfloor 3a+b\rfloor x + b$ are real.  If $\alpha_i$ is a nonreal root of a quadratic with real coefficients, its complex conjugate is also a root.
From this, it follows that $g(x)=4x^2-2\lfloor 3a+b\rfloor x + b$ is some multiple of $f(x)=x^2-3x+4$.  Looking at the $x^2$ term, we find then that $4f(x)=g(x)$.  Equating the coefficients together, we have then $b=4\cdot 4 = 16$ and $-2\lfloor 3a+b\rfloor = -3\cdot 4$ from which it follows after some arithmetic that $\lfloor 3a\rfloor = -10$.
Thus, the possible pairs of points $(a,b)$ are $\{(a,b)~:~\frac{-10}{3}\leq a < -3, b=16\}$
A: I would rearrange the first equation to mak $x^2=3x-4$ and substitute that into the second equation as a starting point. Everything is linear  after that.
