How do you solve # (x^2+2x)^2-2(x^2+2x)-3=0#?

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Answer 1 · 2016-07-14T00:20:59

#x = {-3,-1,1}#

First making #y = x^2+2x# and substituting

#y^2-2y-3=0#

Solving for #y# gives #y = {-1,3}#

The next step is to solve

1) #x^2+2x = -1# giving two equal roots #x = {-1,-1}#
2) #x^2+2x = 3# giving #x = {-3, 1}#

The solutions are

#x = {-3,-1,1}#

also we have

#(x^2 + 2 x)^2 - 2 (x^2 + 2 x) - 3=(x+3)(x+1)^2(x-1)#