How do you solve # 2x^2 + 12x − 14 = 0# by completing the square?

1 Answer
Feb 12, 2017

Answer:

#x=-7color(white)("XX")orcolor(white)("XX")x=+1#

Explanation:

#2x^2+12x-14=0#

#rarr2x^2+12x=14#

#rarr x^2+6x=7#

Now to "complete the square":
If #x^2+6x# are the first two terms of the expansion of a squared binomial: #(x+a)^2 = x^2+2ax+a^2#
then #2ax# must equal #6x#;
that is #a=3# and #a^2=9#

To "complete the square" we must add #color(magenta)(9)# to the expression,
but we can only legally do this if we add #color(magenta)9# to both sides of the equation:
#color(white)("XXX")x^2+6xcolor(magenta)(+9) = 7color(magenta)(+9)#

#color(white)("XXX")rarr (x+3)^2=16#

#color(white)("XXX")rarr x+3=+-4#

#color(white)("XXX")rarr x=-3+-4#

#color(white)("XXX")#which can be written as: # x=-7color(white)("X")orcolor(white)("X")x=1#