Finding and b in an equation with a floor function \begin{align}
    a^2 + ba + c &= 0 & \text{{No real roots.}}\\
    \lfloor a^2 \rfloor + ba + c &= 0 & \text{{At least one real roots.}}
\end{align}
Are there any values of b and c that will make the given number of roots correct?
The first thing I thought about doing was finding the discriminant .In order for the first equation to have no real roots, we must have $b^2 – 4c < 0$. That means that $b^2 < 4c$. 
I know that the discriminant of the second one must be 0 but I am not sure how to express it because a floor function is involved. What should I do? 
Will this approach get me anywhere or are there any better methods? 
 A: \begin{align}
    x^2 + bx + c &= 0 & \text{{No real roots.}}\\
    \lfloor x^2 \rfloor + bx + c &= 0 & \text{{At least one real root.}}
\end{align}
Let's start by noting that the discriminant of $x^2 + bx + c$ is
$b^2 - 4c$. Hence  $x^2 + bx + c$ has no real roots if and only if $b^2 - 4c < 0$, which is true if and only if $c - \dfrac{b^2}{4} > 0$.  For convenience, let's define $$\Delta = c - \dfrac{b^2}{4}$$
and from here on, it will turn out to be necessary to assume that $0 < \Delta < 1$.
Let $f(x) = \lfloor x^2 \rfloor + bx + c$
If $x$ is an integer, then $f(x) = x^2 + bx + c$ which we assume has no real roots. So the roots of $f(x)$ cannot be an integer.
Since $x^2 - 1 < \lfloor x^2 \rfloor \le x^2$, then
$x^2 + bx + (c-1) < f(x) < x^2 + bx + c$
So, if $f(x) = 0$,
\begin{array}{c}
    x^2 + bx + (c-1) < 0 < x^2 + bx + c \\
    -x^2 - bx - c < 0 < -x^2 - bx + (1-c) \\
    0 < x^2 + bx + c < 1 \\
    0 < \left(x + \dfrac b2 \right)^2 + \Delta < 1 \\
    0 \le \left(x + \dfrac b2 \right)^2 < 1 - \Delta \\
    \left|x + \dfrac b2 \right| \lt \sqrt{1 - \Delta}
\end{array}
So all possible solutions to $f(x) = 0$ are in the interval
$\left( -\dfrac b2 - \sqrt{1 - \Delta}, -\dfrac b2 + \sqrt{1 - \Delta} \right)$.
This range can be partitioned into intervals of the form $(-\sqrt{n+1}, -\sqrt n]$ or $[\sqrt n, \sqrt{n+1})$ and each interval will contain $0$ or $1$ solutions as follows.
If $-\dfrac{n+c}{b}\in (\sqrt{n+1}, \sqrt n]$ then it is a root of $f(x)$.
If $-\dfrac{n+c}{b}\in [-\sqrt n, -\sqrt{n-1})$ then it is a root of $f(x)$.
Example
Let $f(x) = \lfloor x^2 \rfloor + 3x + 3$
Then $0 < \Delta = \dfrac 34 < 1$ and $\sqrt{1 - \Delta} = \dfrac 12$
So all possible solutions to $f(x) = 0$ are in the interval
$\left( -\dfrac 32 - \dfrac 12, -\dfrac 32 + \dfrac 12 \right) = ( -2, -1)$.
\begin{array}{|c|c|c|c|}
\hline
\text{n} & \text{interval} & a=-\dfrac{n+c}{b} & \lfloor a^2 \rfloor + ba + c \\
\hline
4 & (-\sqrt 5, -\sqrt 4] & -\dfrac 73 & \text{not in interval} \\
3 & (-\sqrt 4, -\sqrt 3] &         -2 & \text{not in interval} \\
2 & (-\sqrt 3, -\sqrt 2] & -\dfrac 53 &          2 - 5 + 3 = 0 \\
1 & (-\sqrt 2, -\sqrt 1] & -\dfrac 43 &          1 - 4 + 3 = 0 \\
0 & (-\sqrt 1, -\sqrt 0] &         -1 & \text{not in interval} \\
\hline
\end{array}

A: Ok with finding the discriminant.
Now plot $$
\eqalign{
  & y =  - bx - c  \cr 
  & y = x^{\,2}   \cr 
  & y = \left\lfloor {x^{\,2} } \right\rfloor  \cr} 
$$
and that will show you how to proceed.
