When are the roots of x^2 + ax + b complex? What does the sketch look like?

Question

I've Already Answered the First Part Of the Question.

In this question I have already answered the "when are the roots complex part."
When I take the discriminant out of the quadratic formula and plug the formula in; I find that all discriminant results that are less than 0 result in a complex number.

But I am at a loss of how to sketch this?

Is it an odd circle shape on the complex plane or something?

Answer 1

You see how the the x is powered to the second? It means that one side of the parabola is a mirror image to the other side.

It will always be in a u shape line or a c shape

Answer 2

x^2 + ax + b
discriminant: a^2-4(1)(b)<0 then the roots are complex

get the x any y-intercepts of the equation

y=x^2+ax+b
y-intercept
y=b
x-intercept
x^2+ax+b=0
x^2+ax=-b
x^2+ax+a^2/4=-b+a^2/4
(x+a/2)^2=(-4b+a^2)/4
x+a/2=+/-(-4b+a^2)/2
x=+/-(-4b+a^2)-a/2

Answer 3

It has complex roots when a^2 - 4b is negative.

a^2 - 4b < 0
4b > a^2
b> a^2

This is what they want you to graph.
Instead of x and y axes you make them a and b axes

Answer 4

It is a parabola that does not cross the x axis, It is graphed in the normal xy plane. (Real Plane)

Answer 5

i might have to use an Argand Diagram, im not really sure