If #v=u^2-u#, what is #u# in terms of #v#?

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Answer 1 · 2018-01-28T18:30:04

#u = 1/2+-sqrt(v+1/4)#

Given:

#v = u^2-u = (u-1/2)^2-1/4#

Add #1/4# to both ends to get:

#v+1/4 = (u-1/2)^2#

Hence:

#u-1/2 = +-sqrt(v+1/4)#

Then adding #1/2# to both sides we find:

#u = 1/2+-sqrt(v+1/4)#

So for all values of #v# apart from #v=-1/4#, there are two values of #u# for each value of #v#.