How do you solve #x^2 + 6x +8= 0 # by completing the square?

1 Answer
Feb 4, 2016

Answer:

#x=-4 or x=-2#

(see below for method)

Explanation:

Given:
#color(white)("XXX")x^2+6x+8=0#

It will be convenient (although not necessary) to move the constant to the right side as:
#color(white)("XXX")color(blue)(x^2+6x=-8)#

Completing the square

For the general squared binomial: #(x+a)^2 = x^2+2a+a^2#

So if #x^2+6x#
#color(white)("XXX")#are the first two terms of an expanded squared binomial
then #a=3 and a^2=9#
and we need to add #9# (to both sides) to complete the square

#color(blue)(x^2+6x)color(red)(+9) = color(blue)(-8)color(red)(+9)#

#color(green)((x+3)^2= 1)#

If #(x+3)^2=1#
then
#color(white)("XXX")x+3=+-1#

#color(white)("XXX")x=-3+-1#

#color(white)("XXX")x=-4 or x=-2#