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

Question

6659 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-03T16:24:04

$3(x+1/2)^2 + y = 3/4$

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

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

$(3x^2 + 3x) + y = 0$

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

Step 2. Complete the squares for $x$ .

$3x^2 +3x$

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

$3(x^2 +x)$

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

$(1)^2/4 = 1/4$

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

$3(x^2 +x + 1/4 -1/4) = 3(x^2+x+1/4) -3/4 = 3(x+1/2)^2 -3/4$

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

(Nothing to do here.)

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

$3(x+1/2)^2 -3/4 +y = 0$

$3(x+1/2)^2 + y = 3/4$

Check:

$3(x+1/2)^2 -3/4 +y = 3(x^2 +x + 1/4) -3/4 +y$

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

How do you complete the square to solve 3x^2+3x+2y=03x^2+3x+2y=0?