How do you solve #x+1=(x+1)^2#?

1 Answer
sente
Jul 8, 2016

#x in {-1,0}#

Explanation:

We could expand the right hand side, gather all the terms on one side, and then solve the resulting quadratic, however because everything is currently in terms of #x+1#, let's use another approach.

Let #u = x+1#

Then we have

#u = u^2#

#=> u^2-u = 0#

#=> u(u-1) = 0#

#=> u = 0 or u-1 = 0#

#=> x+1 = 0 or x+1-1 = 0#

#:. x = -1 or x = 0#

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