Question 664207: 7+3z^2=4z please help me determine he nature of the solution of this equaion would it be one real, two nonreal or two real???
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 7 + 3z^2 = 4z
arrange is as a quadratic equation
3z^2 - 4z + 7 = 0
:
Using the form ax^2 + bx + c, we have a=3, b=-4, c=7
:
the discriminant determines if the roots are real
d = b^2 - 4*a*c
The rules are:
if d is greater than 0, two real roots
if d is equal to 0 a single root value
if d is less than 0 two complex number solutions (not real)
:
Find d using our equation
d = -4^2 - 4*3*7
d = 16 - 84
d = -16, so two non-real roots
:
:
you can use this to determine the nature of the roots on any quadratic equation
|
|
|
