How do you solve #5x^2 + 20x - 120 = 0# using completing the square?

1 Answer
Jul 3, 2015

Answer:

Logical first step would be to divide everything by #5#

Explanation:

#->x^2+4x-24=0#

Since #(x+a)^2=x^2 +2ax+a^2#
We take half of the coefficient of #x#, and square it:
#x^2+4x+4=(x+2)^2#

We have to balance the #4# we used with the #-24#
So the equation becomes:
#x^2+4x+4-28=0-># add 28:
#(x+2)^2=28#

So #x+2=sqrt28=2sqrt7->x=-2+2sqrt7#
Or #x+2=-sqrt28=-2sqrt7->x=-2-2sqrt7#

(sometimes written as #x_(1,2)=-2+-2sqrt7#)