How do you solve #2x^2 + 8x -3 =0# by completing the square?

2 Answers
Apr 12, 2017

Answer:

#color(green)(x=-2+-sqrt(11/2))#

Explanation:

#2x^2+8x-3=0#

#rArr color(blue)2(x^2+4x)=3#

If #color(orange)(4x)# is the middle term of a squared binomial with the form
#color(white)("XXX")(x+a)^2 = (x^2+color(orange)(2ax)+a^2)#
then
#color(white)("XXX")color(orange)a=color(orange)2#
and
#color(white)("XXX")color(green)(a^2)=color(green)(4)# must be added to #x^2+4x# to complete the square.

This will result in adding #color(blue)2 xxcolor(green)4# to the left side of the equation;
so we must add this to the right side as well to keep the equation valid.

#color(white)("XXX")color(blue)2(x^2+4x+color(green)4)=3+color(blue)2xxcolor(green)4#

Rewriting the parenthesized factor as a squared binomial and simplifying.

#color(white)("XXX")2(x+2)^2=11#

Now divide both sides by #2#, leaving only a squared binomial on the left:
#color(white)("XXX")(x+2)^2=11/2#

Take the square root of both sides (don't forget that there will be both a positive and negative root on the right side):
#color(white)("XXX")x+2=+-sqrt(11/2)#

Subtract #2# from both sides to isolate the variable #x#
#color(white)("XXX")x=-2+-sqrt(11/2)#

Apr 12, 2017

Answer:

# x ~~0.345 or x ~~ -4.345#

Explanation:

# 2x^2 +8x -3 = 0 or 2(x^2+4x) -3 =0 or 2(x^2+4x +4) -3 =8# [Adding 8 on both sides]

or # 2(x+2)^2 =11 or (x+2)^2 = 11/2 or x+2 = +-sqrt(11/2) #

or# x = -2 +- sqrt(11/2) = -2 +- 1/2(sqrt22) :. x ~~0.345 or x ~~ -4.345# [Ans]