# Thread: Factors which include imaginary numbers

1. ## Factors which include imaginary numbers

I am sure this is a very basic procedure, but I am stuck. Can someone please take me through how you determine the solution of the following? Thanks, Frostking

(1 - x) (3 - x) = 1

also

(2 - x) (- 2 - x) = -5

I need these to determine Eigenvalues and I am lost.

2. Originally Posted by Frostking
I am sure this is a very basic procedure, but I am stuck. Can someone please take me through how you determine the solution of the following? Thanks, Frostking

(1 - x) (3 - x) = 1

also

(2 - x) (- 2 - x) = -5

I need these to determine Eigenvalues and I am lost.
3 - 1x - 3x + x^2 = 1
x^2 -4x + 3 = 1
x^2 - 4x + 3 -1 = 0
x^2 - 4x + 2 = 0

Next step
x^2 - 4x + 4 = -2+4
x^2 - 4x + 4 = 2
(x-2)^2 = 2
Square root both sides of the equation
x-2 = +- 1.414
x = 2 +- 1.414
x = 3.414, or x = 0.586

3. ## Factoring with imaginaries

Thanks very much for showing me that I did not have to use i on that one. Could you please tell me how to if I do, say on this one:

I have ( 1 - lambda) (- 3 - lambda) + 5 = 0

So I get -3 + 2 lambda + lambda^2 + 5 = 0

lambda^2 + 2 lambda + 2 = 0

I know from the book that the answer is -1 + i and - 1 - i but I do not know how to get this. Do I just plug values into the quad formula? Thanks for your time. Frostking

4. Originally Posted by Frostking
Thanks very much for showing me that I did not have to use i on that one. Could you please tell me how to if I do, say on this one:

I have ( 1 - lambda) (- 3 - lambda) + 5 = 0

So I get -3 + 2 lambda + lambda^2 + 5 = 0

lambda^2 + 2 lambda + 2 = 0

I know from the book that the answer is -1 + i and - 1 - i but I do not know how to get this. Do I just plug values into the quad formula? Thanks for your time. Frostking
Hi Frostking,

Yes, you just plug them into the quadratic formula.

$\lambda^2+2\lambda+2=0$

$\lambda=\frac{-2\pm\sqrt{2^2-(4)(1)(2)}}{2(1)}$

$\lambda=\frac{-2\pm\sqrt{-4}}{2}$

$\lambda=\frac{-2\pm 2i}{2}$

$\lambda=-1\pm i$