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

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Answer 1 · 2016-05-06T05:11:01

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

Given -

#x^2-6x=13#
Divide the coefficient of #x# by 2 and add it to both sides

#x^2-6x+((-6)/2)^2=13 ++((-6)/2)^2#

You have a perfect square on the left hand side

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

Rewrite it as -

#(x-3)^2=22#

Taking square on both sides, we have

#(x-3)=+-sqrt22#

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