SOLUTION: I have two problems that I am stuck on. 1. 5z^2 - 4z + 1= 0 this is how I started. 5z^2 - 4z = -1 5z^2 4z + 4 = -1 + 4 (z-2)^2 = -3 z-2 = plus or minus i the s

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: I have two problems that I am stuck on. 1. 5z^2 - 4z + 1= 0 this is how I started. 5z^2 - 4z = -1 5z^2 4z + 4 = -1 + 4 (z-2)^2 = -3 z-2 = plus or minus i the s      Log On


   

Question 8591: I have two problems that I am stuck on.
1. 5z^2 - 4z + 1= 0

this is how I started.
5z^2 - 4z = -1
5z^2 4z + 4 = -1 + 4
(z-2)^2 = -3
z-2 = plus or minus i the square root of -3
2 + - i / 5
I dont even know if this is correct.

2. 5t^2 + 4t - 3 = 0

5t^2 + 4t = 3
5t^2 - 4z + 4 = 3 - 4
(t + 2)^2 = -1
t + 2 = =/- -1
then I'm lost. Any help would be greatly appreciated. Thank you.
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
I'm afraid in both exercises, you have tried to solve by completing the square on the z and t terms, but completing the square requires that the coefficient in the squared term MUST BE 1. So completing the square in problems like this really gets ugly. You are better off to use the quadratic formula:

1. 5z%5E2+-4z%2B1+=+0
a=5, b=-4, c=1
z=+%284%2B-sqrt%2816-4%2A5%2A1%29%29%2F%282%2A5%29
z=+%284+%2B-sqrt%28-4%29%29%2F10
z=+%284%2B-i%2A2%29%2F10
z=+%282+%2B-i%29%2F5

2. 5t%5E2+%2B+4t+-+3+=+0
a=5, b=4, c=-3
The result is similar.

R^2 at SCC