SOLUTION: Evaluate the discriminant of each equation. How many real and imaginary solutions does each have? 37. x2 + 5x + 6 = 0 38. 3x2 - 4x + 3 = 0 39. -2x2 - 5x + 4 = 0

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Evaluate the discriminant of each equation. How many real and imaginary solutions does each have? 37. x2 + 5x + 6 = 0 38. 3x2 - 4x + 3 = 0 39. -2x2 - 5x + 4 = 0       Log On


   

Question 177918: Evaluate the discriminant of each equation. How many real and imaginary
solutions does each have?

37. x2 + 5x + 6 = 0
38. 3x2 - 4x + 3 = 0



39. -2x2 - 5x + 4 = 0
40. 16x2 - 8x + 1 = 0
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
D=b%5E2-4ac where D is the discriminant
:
....zero, bē - 4ac = 0 ----->1 real (double)eg (x-3)(x-3)
positive, bē - 4ac > 0 ----->2 real roots
negative, bē - 4ac < 0 ----->2 complex roots
:
Equations taken in the form of ax+by+c=0
:
I will do one and let you complete the rest
:
37. x2 + 5x + 6 = 0
:
a=1,b=5,c=6
:
D=5%5E2-4%28%281%29%286%29%29=25-24=1since D>0 we will have 2 real roots.
38. 3x2 - 4x + 3 = 0
:
%28-4%29%5E2-4%283%29%283%29=16-36=-20D<0
:
so 2 complex roots
:
39. -2x2 - 5x + 4 = 0
:
%28-5%29%5E2-4%28-2%29%284%29=+25%2B32=57D>0 so 2 real roots
:
40. 16x2 - 8x + 1 = 0
:
%28-8%29%5E2-4%2816%29%281%29=64-64=0so 1 real (double) root