How do you solve #x^2+2x-120=0# by completing the square?

1 Answer
Oct 11, 2016

Answer:

#x=10# or #x=-12#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

Use this with #a=(x+1)# and #b=11# as follows:

#0 = x^2+2x-120#

#color(white)(0) = x^2+2x+1-121#

#color(white)(0) = (x+1)^2-11^2#

#color(white)(0) = ((x+1)-11)((x+1)+11)#

#color(white)(0) = (x-10)(x+12)#

Hence roots #x=10# and #x=-12#