I assume you mean "roots of a quadratic equation" since you refer to the "discriminant".
The roots of the equation \(\displaystyle ax^2+ bx+ c= 0\) are \(\displaystyle \dfrac{-b\pm\sqrt{b^2- 4ac}}{2a}\).
Assuming a, b, and c are real numbers, the roots will be real if and only if \(\displaystyle \sqrt{b^2- 4ac}\) is real which is true if and only if \(\displaystyle b^2- 4ac\ge 0\). The roots will be non-real (strictly speaking "complex" includes real). The roots will be "pure imaginary" if and only if \(\displaystyle b^2- 4ac< 0\) and b= 0 which means that \(\displaystyle b^2- 4ac= -4ac< 0\) so that a and c are both positive or both negative.
If a, b, and c are all rational (in particular, if they are integers) then the roots will be rational if and only if \(\displaystyle b^2-4ac\) is rational. If a, b, and c are integers, that will be true if and only if \(\displaystyle b^2- 4ac\) is a perfect square. If they are not integers, then the root will be rational if and only if \(\displaystyle b^2- 4ac\) is a fraction such that both numerator and denominator are perfect squares.
I don't understand what you mean by " These are the roots for practice problems I need help with...1. (-2,0) 2. (-1,0) and (3,0) 3. (1,0) and (6,0)" Do you mean you want to find quadratic equations that have those roots? That has nothing to do with what kind of roots the equation has! Obviously, all of these roots are integers!
In any case, a quadratic equation having roots \(\displaystyle x_0\) and \(\displaystyle x_1\) must be of the form \(\displaystyle a(x- x_0)(x- x_1)= 0\) where a can be any non-zero number. Is that what you wanted?
