# How do you complete the square to solve 3x^2+3x+2y=0?

Jul 3, 2015

$3 {\left(x + \frac{1}{2}\right)}^{2} + y = \frac{3}{4}$

#### Explanation:

$3 {x}^{2} + 3 x + 2 y = 0$

Step 1. Separate the $x$ and $y$ variables.

$\left(3 {x}^{2} + 3 x\right) + y = 0$

Complete the squares for $x$ and $y$ separately.

Step 2. Complete the squares for $x$.

$3 {x}^{2} + 3 x$

(a) Factor out the coefficient of ${x}^{2}$.

$3 \left({x}^{2} + x\right)$

(b) Square the coefficient of $x$ and divide by $4$

${\left(1\right)}^{2} / 4 = \frac{1}{4}$

(c) Add and subtract the result to the term inside parentheses

$3 \left({x}^{2} + x + \frac{1}{4} - \frac{1}{4}\right) = 3 \left({x}^{2} + x + \frac{1}{4}\right) - \frac{3}{4} = 3 {\left(x + \frac{1}{2}\right)}^{2} - \frac{3}{4}$

Step 3. Complete the square for the $y$ term

(Nothing to do here.)

Step 4. Combine the $x$ and $y$ results.

$3 {\left(x + \frac{1}{2}\right)}^{2} - \frac{3}{4} + y = 0$

$3 {\left(x + \frac{1}{2}\right)}^{2} + y = \frac{3}{4}$

Check:

$3 {\left(x + \frac{1}{2}\right)}^{2} - \frac{3}{4} + y = 3 \left({x}^{2} + x + \frac{1}{4}\right) - \frac{3}{4} + y$

= 3x^2 +3x +cancel(3/4) –cancel(3/4) +y = 3x^2 +3x+y