How do you solve #x^2 + 18x + 80 = 0# by completing the square?

2 Answers
Jul 10, 2015

Answer:

I found:
#x_1=-8#
#x_2=-10#

Explanation:

Start by taking #80# to the right and get:
#x^2+18x=-80#
now you need a number that multiplied by #2# gives #18#...could be #9#!
So, add and subtract NOT #9# but #9^2=81#!
#x^2+18xcolor(red)(+81-81)=-80#
take #-81# to the right:
#x^2+18x+81=81-80#
so that you can write:
#(x+9)^2=1#
root square both sides:
#x+9=+-sqrt(1)=+-1#
So you get two solutions:
#x_1=-8#
#x_2=-10#

Jul 10, 2015

Answer:

#x = -8#
or
#x=-10#

Explanation:

Given : #x^2+18x+80=0#

Complete the square:
#color(white)("XXXX")##x^2+18x+(18/2)^2 - (18/2)^2 +80 = 0#
Simplify:
#color(white)("XXXX")##x^2+18x+9^2 -81 + 80 = 0#
Write as a squared binomial:
#color(white)("XXXX")##(x+9)^2 -1 = 0#
Move the constant to the right side
#color(white)("XXXX")##(x+9)^2 = 1#
Take the square root
#color(white)("XXXX")##x+9 = +-1#
Move constant to the right side
#color(white)("XXXX")##x = -10# or #x= -8#