# Complex numbers help needed

1. Nov 18, 2005

### cmantzioros

Complex numbers ... help needed!!

In our exercises we are told to solve for x (element of a complex number)
1. x^2 - 6x + 25=0
The answer is x=3+4i or x=3-4i
Can anyone tell me how these answers were derived??

Last edited: Nov 18, 2005
2. Nov 18, 2005

### NateTG

Are you familiar with the quadratic formula?

3. Nov 18, 2005

### cmantzioros

x^4+3x^2-4=0

or

x^2+ix+6=0

4. Nov 18, 2005

### andrewchang

can't you just use the quadratic equation?

edit:

whoops i was too slow.

5. Nov 18, 2005

### cmantzioros

Quad. form. can only be used when the x-term is of degree 2 ... any idea on how to solve for x in the above eqns?

6. Nov 18, 2005

### andrewchang

is it possible to write it into polar complex form and solve the equation?

7. Nov 18, 2005

### cmantzioros

We haven't covered this topic yet so I'm not quite sure how you would use it

8. Nov 18, 2005

### Hammie

You could always complete the square, and set that equal to -16.

x^2 - 6x + 9 + (25-9) = 0

x^2 - 6x + 9 = -16

can you take it from there?

quadratic formula does basically the same thing, and you can skip half a dozen steps if you remember it..

Last edited: Nov 18, 2005
9. Nov 18, 2005

### NateTG

Those are both quite doable with the quadratic formula:
$$x^4+3x^2-4=\left(x^2\right)^2+3 \left(x^2\right) + 3$$
so
$$x^2= \frac{-3 \pm \sqrt{3^2-4(1)(3)}}{2}$$
and
$$x^2+ix+6$$
is no problem:
$$x=\frac{-i \pm \sqrt{i^2-4(1)(6)}}{2}$$

10. Nov 18, 2005

### cmantzioros

Thanks that does work ... any ideas on the other ones?

11. Nov 18, 2005

### cmantzioros

Thanks for the help. I appreciate it.

12. Nov 18, 2005

### Hammie

x^4 + 3x^2 - 4 = 0 factors into two "pieces". One has a complex solution, the other does not.

Note: there are four roots to the equation. You should have two complex, and two real solutions.

Last edited: Nov 18, 2005
13. Nov 18, 2005

In this case the quadratic formula works, but for other high polynomial funtions you're probably better off with synthetic devision, factoring, and the like. For instance:

(1)^4 + 3(1)^2 - 4 = 0
So:
x^4 + 3x^2 - 4 = (x-1)(x^3 + x^2 + 4x + 4) = 0
(-1)^3 + (-1)^2 + 4(-1) + 4 = 0
So:
(x-1)(x+1)(x^2 + 4) = 0

Real roots are 1 and -1
Complex roots are +2i and -2i