If #(x+1)^2=4# and #(x-1)^2=16#, what is the value of #x#?

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Answer 1 · 2018-05-19T16:20:28

#x=-3#

First solve:

#(x+1)^2 = 4#

#sqrt((x+1)^2) = +-sqrt(4)#

#x + 1 = +-2#

#x = 1, -3#

Now solve:

#(x-1)^2 = 16#

#sqrt((x-1)^2) = +-sqrt(16)#

#x - 1 = +-4#

#x = 5, -3#

The only result that satisfies both equations is #-3#