How do you complete the square for #2x^2 + y^2 + 3x + 4y = 0#?

1 Answer
Jul 2, 2015

Answer:

You complete the square for the #x# and #y# terms separately and get

#2(x+3/4)^2 + (y+2)^2 = 41/8#

Explanation:

#2x^2 +y^2 +3x +4y = 0#

Step 1. Group the #x# and #y# terms separately.

#(2x^2 + 3x )+ (y^2 + 4y) = 0#

Step 2. Complete the square for the #x# terms.

#2x^2 + 3x = 0#

#2(x^2 +3/2x) = 0#

#(3/2)^2/4 = (9/4)/4 = 9/16#

#2(x^2 + 3/2x + 9/16 - 9/16) = 0#

#2((x+3/4)^2 -9/16) = 0#

#2(x+3/4)^2 - 9/8 = 0#

#2(x+3/4)^2 =9/8#

Step 3. Complete the square for the #y# terms.

#y^2 + 4y = 0#

#4^2/4 = 16/4 = 4#

#y^2 + 4y +4 -4= 0#

#(y+2)^2 – 4 = 0#

#(y+2)^2 = 4#

Step 4. Recombine the #x# and #y# terms.

#2(x+3/4)^2 = 9/8#

#(y+2)^2 = 4#

#bar(2(x+3/4)^2 + (y+2)^2 = 9/8 +4#

#2(x+3/4)^2 + (y+2)^2 = 9/8 +32/8#

#2(x+3/4)^2 + (y+2)^2 = 41/8#

Check:

#2(x+3/4)^2 + (y+2)^2 -41/8#

#= 2(x^2 + 3/2x + 9/16) + (y^2 +2y+4) - 41/8#

#= 2x^2 +3x +9/8 +y^2 + 2x +4 -41/8#

#= 2x^2 +y^2+3x + 2y +9/8 +4 - 41/8#

#2x^2 +y^2+3x + 2y + 41/8 -41/8 =0#

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