How would I solve this over the set of complex numbers?

Question

m^2+12m+5=0

? I did it once but I guess I did it wrong and I can't figure out the right way?!

Answer 1

use the quadratic equation
-b/2a + or - (the square root of b^2-4ac)/2a
-12/2 + or - (the square root of 12^2 - (4x5))/2

so use a calculator

Answer 2

Use the quadratic formula.

But, let's check to see if the roots are complex by checking the discriminant:

b2-4ac = 12^2 - 4*1*5
= 144 -20 = 124

this is greater than zero indicating that the two roots are real!

Just search for 'quadratic formula' is you don't remember or don't have a book handy.

+ add
The solutions are not very pretty. They are:

-12/2 + sqrt(31) = -0.4322
-12/2 - sqrt(31) = -11.57

The expressions with the square roots are exact, and irrational. The decimal values are approximate.

Answer 3

m = [ - 12 ± √(144 - 20) ] / 2
m = [- 12 ± √(124)] / 2
m = [- 12 ± 2 √31 ] / 2
m = - 6 ± √31 (no complex numbers)

Answer 4

it is not factorable, so use quadratic formula
[ -12 +/- sq (12^2 - 4(1)(5) ) ] / 2 =
[ -12 +/- sq (144 - 20) ] / 2 =
[ -12 +/- sq (124) ] / 2 =
[ -12 +/- 2 sq(31) ] / 2 =
-6 +/- sq(31)