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

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Answer 1 · 2015-07-19T21:32:35

#x=3+sqrt13, 3-sqrt13#

#6x+4=x^2#

Completing the square means forcing a perfect square trinomial on the left side of the equation.

Bring all terms to the left side.

#-x^2+6x+4=0#

Multiply boths sides times #-1#.

#-1(-x^2+6x+4=0)# =

#x^2-6x-4=0#

Add #4# to both sides of the equation.

#x^2-6x=4#

Divide the x-term by 2 and square the result. Add 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 on the left side. #a^2-2ab+b^2=(a-b)^2#, where #a=x and b=3#.

Substitute #(x-3)^2# into the equation.

#(x-3)^2=13# =

Take the square root of both sides and solve for #x#.

#x-3=+-sqrt13#

Add #3# to both sides.

#x=3+-sqrt13#

Solve for #x#.

#x=3+sqrt13#

#x=3-sqrt13#