SOLUTION: solve over the set of complex numbers (x+5)^2=-36 (given) if I work it out I get x^2-10x+61=0 but now what, it won't foil, x^2-10x+25=-36 x^2-10x+25= -61+25 completing t

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: solve over the set of complex numbers (x+5)^2=-36 (given) if I work it out I get x^2-10x+61=0 but now what, it won't foil, x^2-10x+25=-36 x^2-10x+25= -61+25 completing t      Log On


   

Question 695868: solve over the set of complex numbers
(x+5)^2=-36 (given)
if I work it out I get x^2-10x+61=0
but now what, it won't foil,
x^2-10x+25=-36
x^2-10x+25= -61+25 completing the square gets me right back where I started...
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
To solve a quadratic equation for complex solutions, you work just as you do for real solutions,
except that you go further than the point where you would say the equation has no real solutions,
because a negative number cannot be the square of a real number,
but it is the square of a complex number.

%28x%2B5%29%5E2=-36 <--> %28x%2B5%29%5E2=%28-1%29%286%5E2%29
The only real (and complex) numbers that squared equal 36=6%5E2 are -6 and 6
The only complex numbers that squared equal -1 are -i and i
So, the only complex numbers that squared equal -36=%28-1%29%286%5E2%29 are -6i and 6i.
So, either x%2B5=-6i <--> highlight%28x=-5-6i%29,
or x%2B5=6i <--> highlight%28x=-5%2B6i%29