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

Jul 2, 2015

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

$2 {\left(x + \frac{3}{4}\right)}^{2} + {\left(y + 2\right)}^{2} = \frac{41}{8}$

Explanation:

$2 {x}^{2} + {y}^{2} + 3 x + 4 y = 0$

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

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

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

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

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

${\left(\frac{3}{2}\right)}^{2} / 4 = \frac{\frac{9}{4}}{4} = \frac{9}{16}$

$2 \left({x}^{2} + \frac{3}{2} x + \frac{9}{16} - \frac{9}{16}\right) = 0$

$2 \left({\left(x + \frac{3}{4}\right)}^{2} - \frac{9}{16}\right) = 0$

$2 {\left(x + \frac{3}{4}\right)}^{2} - \frac{9}{8} = 0$

$2 {\left(x + \frac{3}{4}\right)}^{2} = \frac{9}{8}$

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

${y}^{2} + 4 y = 0$

${4}^{2} / 4 = \frac{16}{4} = 4$

${y}^{2} + 4 y + 4 - 4 = 0$

(y+2)^2 – 4 = 0

${\left(y + 2\right)}^{2} = 4$

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

$2 {\left(x + \frac{3}{4}\right)}^{2} = \frac{9}{8}$

${\left(y + 2\right)}^{2} = 4$

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

$2 {\left(x + \frac{3}{4}\right)}^{2} + {\left(y + 2\right)}^{2} = \frac{9}{8} + \frac{32}{8}$

$2 {\left(x + \frac{3}{4}\right)}^{2} + {\left(y + 2\right)}^{2} = \frac{41}{8}$

Check:

$2 {\left(x + \frac{3}{4}\right)}^{2} + {\left(y + 2\right)}^{2} - \frac{41}{8}$

$= 2 \left({x}^{2} + \frac{3}{2} x + \frac{9}{16}\right) + \left({y}^{2} + 2 y + 4\right) - \frac{41}{8}$

$= 2 {x}^{2} + 3 x + \frac{9}{8} + {y}^{2} + 2 x + 4 - \frac{41}{8}$

$= 2 {x}^{2} + {y}^{2} + 3 x + 2 y + \frac{9}{8} + 4 - \frac{41}{8}$

$2 {x}^{2} + {y}^{2} + 3 x + 2 y + \frac{41}{8} - \frac{41}{8} = 0$

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