How do you complete the square to solve 3x^2+3x+2y=03x2+3x+2y=0?

1 Answer
Ernest Z.
Jul 3, 2015

3(x+1/2)^2 + y = 3/43(x+12)2+y=34

Explanation:

3x^2 + 3x + 2y = 03x2+3x+2y=0

Step 1. Separate the xx and yy variables.

(3x^2 + 3x) + y = 0(3x2+3x)+y=0

Complete the squares for xx and yy separately.

Step 2. Complete the squares for xx.

3x^2 +3x3x2+3x

(a) Factor out the coefficient of x^2x2.

3(x^2 +x)3(x2+x)

(b) Square the coefficient of xx and divide by 44

(1)^2/4 = 1/4(1)24=14

(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 -3/43(x2+x+14−14)=3(x2+x+14)−34=3(x+12)2−34

Step 3. Complete the square for the yy term

(Nothing to do here.)

Step 4. Combine the xx and yy results.

3(x+1/2)^2 -3/4 +y = 03(x+12)2−34+y=0

3(x+1/2)^2 + y = 3/43(x+12)2+y=34

Check:

3(x+1/2)^2 -3/4 +y = 3(x^2 +x + 1/4) -3/4 +y3(x+12)2−34+y=3(x2+x+14)−34+y

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

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