How do you solve #x^2+6x+13=0# by completing the square?

1 Answer
Oct 30, 2016

Answer:

#x=-3+-2i#

Explanation:

Solve by completing the square.

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

Move the constant to the right side by subtracting 13 from each side.

#x^2+6xcolor(white)(aaaa)=-13#

Divide the coefficient of the #x# term by #2#.

#x^2+color(red)6xcolor(white)(aaaa)=-13#

#color(red)6/2=color(blue)3#

Square the result and add it to both sides.

#color(blue)3^2=color(magenta)9#

#x^2+6x +color(magenta)9=-13+color(magenta)9#

Factor the left side and simplify the right side.

#(x+color(blue)3)(x+color(blue)3)=-4#

Rewrite as the square of the binomial. Note that the #color(blue)3# in the binomial is the same value #color(blue)3# that resulted from dividing the coefficient of the #x# term by #2#.

#(x+color(blue)3)^2=-4#

Square root both sides and solve for #x#.

#sqrt((x+color(blue)3)^2)=sqrt(-4)#

#x+color(blue)3=+-2i#

#x=-3+-2i#