Question 411675: How do you solve this equation over the complex numbers?
2t^2-2sqrt( 2 ) t+1=0
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 2t^2 - 2sqrt(2)t + 1 = 0
lets look at the discriminant of this equation, where a=2, b=2sqrt(2), c=1
[2sqrt(2)]^2 - 4 * 2 * 1 =
8 - 8 = 0; it has one real root
:
Try completing the square
t^2 should have a coefficient of 1, divide eq by 2
t^2 - sqrt(2)t + .5 = 0
t^2 - sqrt(2)t + ___ = -.5
Divide the coefficient of t by 2 and square it: (.5sqrt(2))^2 = .25(2) = .5
t^2 - sqrt(2)t + .5 = -.5 + .5
t^2 - sqrt(2)t .5 = 0
(t - .5sqrt(2)^2 = 0
same as
t - .5sqrt(2) = 0
t = +.5sqrt(2) a single root (~ +.707)
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