How do you solve #2w ^ { 2} + 4w + 11= ( w - 2) ^ { 2}#?
1 Answer
Apr 13, 2017
Explanation:
-
Firstly expand the right side:
#(w-2)^2 = (w-2)(w-2)=w^2-4w+4# -
subtract
#w^2-4w+4# from both sides:
#w^2+8w+7=0# -
factor the left side
#(w+7)(w+1)=0# -
By the zero product property
#(w+7) = 0 or (w+1) = 0#
graph{y=(x+7)(x+1) [-20, 20, -10, 10]}