How do you solve #x^2 - 3x +2 = 0# by completing the square?

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Answer 1 · 2016-07-04T19:04:13

#x = 2 or x = 1#

#ax^2 + bx +c =0# is the general form of a quadratic trinomial.

#x^2 - 3x " " = -2 " move the constant to the RHS"#

Add #color(red)((b/2)^2)# to both sides to form the square of a binomial. This step is the completing of the square.

#x^2 - 3x + color(red)((3/2)^2) " = -2 + color(red)((3/2)^2)#

Add #(b/2)^2# to both sides to form the square of a binomial.

#(x - 3/2)^2 = -2 + 9/4 = 1/4#

#x -3/2 = +-sqrt(1/4)#

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

#x = +1/2 +1 1/2 " or " x = -1/2 +1 1/2#

#x = 2 or x = 1#