How do you solve using the completing the square method #x^2-5x+3=0#?

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Answer 1 · 2017-11-04T02:48:29

#x=1/2(5+-sqrt13)#
[#x approx 4.30278 or approx 0.69722#]

#x^2-5x+3=0#

#x^2-5x =-3#

Our objective here is to make the LHS into a perfect square.

Add #(5/2)^2# to both sides:

#x^2-5x +(5/2)^2=-3+(5/2)^2#

Notice that the LHS #= (x-5/2)^2#

Hence, #(x-5/2)^2 = -3 +25/4#

#(x-5/2)^2 = (-12 +25)/4 =13/4#

#:. x-5/2 =+-sqrt(13/4)#

#x= 5/2+-sqrt13/sqrt4#

#= 1/2(5+-sqrt13)#

#x approx 4.30278 or approx 0.69722#