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

Oct 24, 2017

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

$x = - \frac{1}{2} \pm \sqrt{\frac{1}{4} - \frac{2}{3} y}$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} 3 {x}^{2} + 3 x + 2 y = 0$

Remember that the "target" completed square must be of the form:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b} \left(= 0\right)$

First we will extract the $\textcolor{g r e e n}{m}$ factor from the first 2 terms:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{3} \left({x}^{2} + 1 x\right) + 2 y = 0$

We also need to remember that the first 2 terms of the expansion of a squared binomial ${\left(x - a\right)}^{2}$ are ${x}^{2}$ and $- 2 a x$ (and that the third term is ${a}^{2}$)

If $\left({x}^{2} + 1 x\right)$ are the first 2 terms of an expanded squared binomial then $\textcolor{red}{a} = - \frac{1}{2}$ and ${a}^{2} = \frac{1}{4}$ will need to be added to these 2 terms to give an expanded completed square.
So this potion of our solution must look like $\left({x}^{2} + 1 x \textcolor{m a \ge n t a}{+ \frac{1}{4}}\right)$

Notice that this new term $\textcolor{m a \ge n t a}{\frac{1}{4}}$ is actually being multiplied by the $\textcolor{g r e e n}{m} = \textcolor{g r e e n}{3}$ factor;
so we are actually adding $\textcolor{g r e e n}{3} \times \textcolor{m a \ge n t a}{\frac{1}{4}} = \frac{3}{4}$

To avoid changing the value of the expression if we are going to add $\frac{3}{4}$ we will also need to subtract $\frac{3}{4}$;
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
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{3} {\left(x + \frac{1}{2}\right)}^{2} + \textcolor{b l u e}{2 y - \frac{3}{4}} = 0$
or in proper vertex form
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{3} {\left(x - \textcolor{red}{\left(- \frac{1}{2}\right)}\right)}^{2} + \textcolor{b l u e}{2 y - \frac{3}{4}} = 0$

This ends the "completing the square portion";
now onto the solution:
$\textcolor{w h i t e}{\text{XXX}} 3 {\left(x + \frac{1}{2}\right)}^{2} = \frac{3}{4} - 2 y$

$\textcolor{w h i t e}{\text{XXX}} {\left(x + \frac{1}{2}\right)}^{2} = \frac{1}{4} - \frac{2}{3} y$

$\textcolor{w h i t e}{\text{XXX}} x + \frac{1}{2} = \pm \sqrt{\frac{1}{4} - \frac{2}{3} y}$

$\textcolor{w h i t e}{\text{XXX}} x = - \frac{1}{2} \pm \sqrt{\frac{1}{4} - \frac{2}{3} y}$