SOLUTION: Solve quadratic equation: x^2+3x+5=0 a. -3 plus minus i square root 29 over 2 b. 3 plus minus square root 7 over 2 c. 5 plus minus square root 13 over 2 d. -3 plus min

Algebra ->  Radicals -> SOLUTION: Solve quadratic equation: x^2+3x+5=0 a. -3 plus minus i square root 29 over 2 b. 3 plus minus square root 7 over 2 c. 5 plus minus square root 13 over 2 d. -3 plus min      Log On


   

Question 136932: Solve quadratic equation:
x^2+3x+5=0



a. -3 plus minus i square root 29 over 2
b. 3 plus minus square root 7 over 2
c. 5 plus minus square root 13 over 2
d. -3 plus minus i square root 11 over 2
Answer by MORUPHOSUOLALE(24) About Me  (Show Source):
You can put this solution on YOUR website!
Since this cannot be slove by factorization.
I will suggest using completing the square method or the Almight Fomulae.
Using Almighty fomulae-
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
In your equation a= 1 , b=3 and c=5
Therefore- x+=+%28-3+%2B-+sqrt%28+3%5E2-4%2A1%2A5+%29%29%2F%282%2A1%29+
x+=+%28-3+%2B-+sqrt%28+9-20+%29%29%2F%282%29+


x+=+%28-3+%2B-+sqrt%28+-11+%29%29%2F%282%29+
However, since the negative 11 is in the square root.This tells us it's a complex number.
We have to apply the complex numbers theorem that states that i^2= -1
Therefore i=sqr (-1)
This tells us that option D is your correct answer.
Hope it helps u.

Moruph
mosuolal@uncc.edu