If #v=u^2-u#, what is #u# in terms of #v#?
1 Answer
Jan 28, 2018
Explanation:
Given:
#v = u^2-u = (u-1/2)^2-1/4#
Add
#v+1/4 = (u-1/2)^2#
Hence:
#u-1/2 = +-sqrt(v+1/4)#
Then adding
#u = 1/2+-sqrt(v+1/4)#
So for all values of