Discover How to Solve Complex Numbers with Ease

[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??

 

[NateTG]

Can anyone tell me how these answers were derived??[/B]

Are you familiar with the quadratic formula?

 

[cmantzioros]

Yes, I may be able to get the answer using the quad. formula for this one but what about if I had:

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

or

x^2+ix+6=0

 

[andrewchang]

can't you just use the quadratic equation?

edit:

whoops i was too slow.

 

[cmantzioros]

can't you just use the quadratic equation?

edit:

whoops i was too slow.

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?

 

[andrewchang]

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

 

[cmantzioros]

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

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

 

[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..

 

[NateTG]

Yes, I may be able to get the answer using the quad. formula for this one but what about if I had:
x^4+3x^2-4=0
or
x^2+ix+6=0

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:
$$x=\frac{-i \pm \sqrt{i^2-4(1)(6)}}{2}$$

 

[cmantzioros]

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..

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

 

[cmantzioros]

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:
$$x=\frac{-i \pm \sqrt{i^2-4(1)(6)}}{2}$$

Thanks for the help. I appreciate it.

 

[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.

 

[Quadratic]

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.

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