How do you solve using the completing the square method #3x^2 +3x +2y=0#?

1 Answer
Oct 24, 2017

Answer:

Note that since you have a single equation in two variables, the "solution" can only be written with the inclusion of a variable.

#x=-1/2+-sqrt(1/4-2/3y)#

Explanation:

Given
#color(white)("XXX")3x^2+3x+2y=0#

Remember that the "target" completed square must be of the form:
#color(white)("XXX")color(green)m(x-color(red)a)^2+color(blue)b(=0)#

First we will extract the #color(green)m# factor from the first 2 terms:
#color(white)("XXX")color(green)3(x^2+1x)+2y=0#

We also need to remember that the first 2 terms of the expansion of a squared binomial #(x-a)^2# are #x^2# and #-2ax# (and that the third term is #a^2#)

If #(x^2+1x)# are the first 2 terms of an expanded squared binomial then #color(red)a=-1/2# and #a^2=1/4# will need to be added to these 2 terms to give an expanded completed square.
So this potion of our solution must look like #(x^2+1xcolor(magenta)(+1/4))#

Notice that this new term #color(magenta)(1/4)# is actually being multiplied by the #color(green)m=color(green)3# factor;
so we are actually adding #color(green)3xxcolor(magenta)(1/4)=3/4#

To avoid changing the value of the expression if we are going to add #3/4# we will also need to subtract #3/4#;
so we will have
#color(white)("XXX")color(green)3(x^2+1xcolor(magenta)(+1/4))+2ycolor(magenta)(-color(green)3/4=0#

Re-writing with a squared binomial
#color(white)("XXX")color(green)3(x+1/2)^2+color(blue)(2y-3/4)=0#
or in proper vertex form
#color(white)("XXX")color(green)3(x-color(red)((-1/2)))^2+color(blue)(2y-3/4)=0#

This ends the "completing the square portion";
now onto the solution:
#color(white)("XXX")3(x+1/2)^2=3/4-2y#

#color(white)("XXX")(x+1/2)^2=1/4-2/3y#

#color(white)("XXX")x+1/2=+-sqrt(1/4-2/3y)#

#color(white)("XXX")x=-1/2+-sqrt(1/4-2/3y)#