what kind of solution does ax^2 -bx + c = 0 have if b^2 - 4ac < 0
I think an ellipse.. is this correct?
First, that is not a graph.
Did you mean
y=+-sqrt(ax^2+bx+c?) if a>0, it is an ellipse, if a<0 it is a hyperbola.
So othewise, I don't know what you are asking. if b^2-4ac less than zero, then the roots are not in the real domain.
No, the equation ax^2 - bx + c = 0 represents a quadratic equation, not an ellipse. The condition you mentioned, b^2 - 4ac < 0, actually determines the type of solutions that the quadratic equation has.
To determine the nature of the solutions, you can use the discriminant, which is the expression inside the square root of the quadratic formula. In this case, the discriminant is b^2 - 4ac.
If b^2 - 4ac is less than zero (i.e., b^2 - 4ac < 0), then it means the discriminant is negative. This situation leads to the quadratic equation having no real solutions. Instead, it has two complex conjugate solutions, which are in the form of "a + bi" and "a - bi", where "i" represents the imaginary unit.
Therefore, when b^2 - 4ac < 0, the quadratic equation ax^2 - bx + c = 0 will have two complex solutions, rather than real solutions or representing an ellipse.