How do you solve #x^2-6x-4=0# by completing the square?

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Answer 1 · 2015-07-16T18:06:53

#x=3+sqrt13, 3-sqrt13#

#x^x-6x-4=0#

Completing the square means that we will force a perfect square trinomial on the left side of the equation, then solve for #x#.

Add #4# to both sides of the equation.

#x^2-6x=4#

Divide the coefficient of the #x# term by #2#, then square the result. Add it to both sides of the equation.

#((-6)/2)^2=-3^2=9#

#x^2-6x+9=4+9# =

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

We now have a perfect square trinomial in the form #a^2-2ab+b^2=(a-b)^2#, where #a=x and b=3#.

Substitute #(x-3)^2# for #x^2-6x+9#.

#(x-3)^2=13#

Take the square root of both sides.

#x-3=+-sqrt13#

Add #3# to both sides of the equation.

#x=3+-sqrt13#

Solve for #x#.

#x=3+sqrt13#

#x=3-sqrt13#