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

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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#