How do you complete the square for #2x^2 + 8x#?

1 Answer
Jul 15, 2015

Answer:

Extract the coefficient of #x^2# as a factor then add (and ultimately subtract) the square of #1/2# of the coefficient of #x#

Explanation:

Given #2x^2+8x#

Extract the coefficient of #x^2# as a factor
#color(white)("XXXX")##=(2)(x^2+4x)#

Half of the coefficient of #x# is #1/2xx4 = 2#
So the square of half the coefficient of #x# is #2^2 = 4#

Add (and subtract) the square of half the coefficient of #x#
#color(white)("XXXX")##= (2)(x^2+4x+2^2 -4)#

Which could be written as
#color(white)("XXXX")##=(2)((x+2)^2 -4)#

In case you were wondering "why the square of half the coefficient of #x#?)
#color(white)("XXXX")#We are trying for a squared binomial of the form #(x+a)^2#
#color(white)("XXXX")#Since
#color(white)("XXXX")##color(white)("XXXX")##(x+a)^2 = x^2+2ax+a^2#
#color(white)("XXXX")#Given the first 2 terms in the form:
#color(white)("XXXX")##color(white)("XXXX")##x^2+d#
#color(white)("XXXX")##color(white)("XXXX")##color(white)("XXXX")#(#d=2a#)
#color(white)("XXXX")#We need to add #(d/2)^2# to get a "squared form"