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

1 Answer
Apr 1, 2016

Answer:

Using the completing the squares method as requested.
#x=2# or #x=-3#

Explanation:

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

Adding #6# to both sides to clear the constant from the left side
#color(white)("XXX")x^2+x=6#

If #x^2+x# are the first two terms of an expanded squared binomial
#(x+a)^2=x^2+2ax+a^2#
then #2a=1# and #a=1/2#

Therefore, in order to complete the square, we will need to add #a^2=(1/2)^2# to both sides:
#color(white)("XXX")x^2+xcolor(red)(+(1/2)^2)=6color(red)(+(1/2)^2)#

Rewriting as a squared binomial and simplifying the right side
#color(white)("XXX")(x+1/2)^2=6 1/4=25/4#

Taking the square root of both sides:
#color(white)("XXX")x+1/2= +-sqrt(25/4)=+-5/2#

Subtracting #1/2# from both sides
#color(white)("XXX")x=5/2-1/2=2color(white)("XX")orcolor(white)("XX")x=-5/2-1/2=-3#