How do you solve #(c + 1) ^ { 2} - 4( c + 1) + 3= 0#?

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Answer 1 · 2018-06-26T11:30:53

We can view this as a quadratic equation in #c+1# and factor that way:

#0=(c+1)^2-4(c+1)+3=( (c+1)-3)((c+1)-1)=(c-2)c#

#c=2 or c=0#

Answer 2 · 2018-06-26T11:32:39

#c = {0,2} #

Let #c+1 = x #

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

#=> (x-3)(x-1) = 0 #

#=> x = 3 #

#=> x =1 #

Hence

#c + 1 = 3 => c = 2 #

#c+1 = 1 => c = 0 #

#c = { 2,0} #