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

1 Answer
Jul 5, 2015

Answer:

Force a perfect square trinomial on the left side of the equation. Take the square root of both sides and solve for #x#.

Explanation:

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

Force a perfect square trinomial such that #a^2+2ab+b^2=(a+b)^2# on the left side of the equation. Then take the square root of both sides and solve for #x#.

Add #4# to both sides of the equation.

#x^2+6x=4#

Take the coefficient of the #x# term and divide it by #2# and square the result.

#6/2=3;# #3^2=9#

Add #9# to both sides.

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

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

#a=x;# #b=3#

#(x+3)^2=13#

Take the square root of both sides.

#x+3=+-sqrt13#

Subtract #3# from both sides.

#x=-3+sqrt13#

#x=-3-sqrt13#