Question

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

Answer 1

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)`