# Homework Help: How to use a complex number?

1. Apr 29, 2015

### usermath

When I solve a quadratic equation I need to find a Discriminant. If D>0 I have no problem. I can find x1 and x2. And when I draw a parabola I can see the x1 and x2 on a X-line.

But when D<0 I don't understand where I can find x1 and x2 on a plot of function.

For example for 5x2+2x+1=0
I have complex solutions: x1 = -0.2-0.4j and x2 = -0.2+0.4j

Ok. It's the answer, but what I can do with one? Where can I find this points on the plot of function?

It's a plot of funciton y=5x2+2x+1

Coordinates of a Minimum Point is X = -0,2 Y = 0,8.

What does (y=0; x=-0.2+0.4) mean? How can I use this expression? Where I can find the point with coordinates y=0; x=-0.2+0.4j ?

2. Apr 29, 2015

### SteamKing

Staff Emeritus
Not on the Real Plane, unfortunately.

Since the parabola shown does not intersect the x-axis, it has no real roots.

Because the discriminant D < 0, it does have a pair of complex conjugate roots which satisfy the equation.

Applying the Quadratic formula should give you:

x = {-2 ± √[(-4)2-4(5)(1)]} / (2)(5)

or x = -0.2 ± 0.2 i [Note: you made an arithmetic error when you took the square root of D]

3. Apr 29, 2015

### paisiello2

I think traditionally "i" is used instead of "j" for complex numbers but that doesn't matter for your question.

The graph you are showing is a 2-dimensional graph of the x-y plane. x and y are real numbers so this is the "real" plane. Graphically there are no real values for x where y=0.

However, if you were to create another plane perpendicular to the x-y plane where you now use j as the "z"-axis then this is the "complex" plane. If you now plot the values of y for different values of x+zj you would get something in 3-dimensions that looks like a series of inverted parabolas that look like a saddle (called a hyperbolic paraboloid). Graphically you would see that the saddle intersects the y=0 plane on two lines: x = -0.2+0.2zj and x = -0.2 - 0.2zj.

In the picture z has taken the place of y and vice versa.

4. Apr 30, 2015

### usermath

Thanks a lot. It was very useful answers for me.