How do you solve # x^2 - 6x =13# by completing the square?

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Answer 1 · 2018-03-24T05:36:03

#x=3+sqrt22#
#x=3-sqrt22#

Given -

#x^2-6x=13#

#x^2-6x+9=13+9#

#(x-3)^2=22#

#x-3=+-sqrt22#

#x=+-sqrt22 +3#

#x=3+sqrt22#
#x=3-sqrt22#

Answer 2 · 2018-03-24T05:44:30

#x= sqrt22+3 or -sqrt22+3#

#x^2 -6x =13#

#x^2 -6x -13=0#

Multiply the equation by 4

#4(x^2 -6x -13=0)#

#4x^2 -24x-52=0#

#(2x)^2 -24x-52=0#

#(2x)^2-2×(2x)×(6)+6^2-6^2-52=0#

#(2x-6)^2 -36-52 =0#

#(2x-6)^2 -88=0#

#(2x-6)^2=88#

#2x-6= +-sqrt 88#

#2x-6=sqrt 88 or 2x-6=-sqrt88#

#2x-6=2sqrt22 or 2x-6=-2sqrt22#

#x= sqrt22+3 or x= -sqrt22+3#