How do you solve the equation #x^2-4x+5=0# by completing the square?

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Answer 1 · 2017-03-06T16:07:21

#x=2+-i#

#x^2-4x+5=0#

1) half the coefficient of #x# square, add and subtract to balance

#color(blue)(x^2-4x+2^2)+5-2^2=0#

#"2) the blue terms form a perfect square"#

#color(blue)((x-2)^2)+5-4=0#

#(x-2)^2+1=0#

3) solve for #x#

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

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

#x-2=+-i#

#:.x=2+-i#