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

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Answer 1 · 2018-06-25T19:56:28

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Expand and rearrange:

#4x^2 +4x + 1 = x+2 #

#=> 4x^2 +3x -1 = 0 #

Factor:

#(4x-1)(x+1) = 0 #

#4x -1 = 0 => x = 1/4 #

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

#x = {-1,1/4 } #