2x^2 + 2y^2 + x + y = 0

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- Jun 17th 2009, 07:49 AMJoker37Sketch the graph of the following:
2x^2 + 2y^2 + x + y = 0

Show complete working out. - Jun 17th 2009, 08:30 AMHallsofIvy
Your question is not at all clear. You

**could**"sketch" the graph by calculating several points: if x= 0, $\displaystyle 2y^2+ y= y(2y+1)= 0$ so y= 0 or -1/2. $\displaystyle (0, 0)$ and $\displaystyle (0, -\frac{1}{2})$ are points on the graph. Taking y= 0, we get, in exactly the same way, $\displaystyle (0,0)$ which we had before, $\displaystyle (0,\frac{1}{2})$. Get enough points that way and you can sketch the graph.

But I suspect that you are intended to**complete the square**to see what kind of figure this is:

If you divide through by 2 you get $\displaystyle x^2+ \frac{1}{2}x+ y^2+ \frac{1}{2}y= 0$. Presumably you know that $\displaystyle (x+ a)^2= x^2+ 2ax+ a^2$. Comparing that to $\displaystyle x^2+ \frac{1}{2}x$, they match if a= 1/4 and we can make this a "perfect square by adding $\displaystyle \left(1/4\right)^2= 1/16$. Doing that for both x and y, $\displaystyle x^2+ \frac{1}{2}x+ \frac{1}{4}+ y^2+ \frac{1}{2}y+ \frac{1}{4}= \frac{1}{4}+ \frac{1}{4}$ or $\displaystyle (x+\frac{1}{2})^2+ (y+\frac{1}{2})^2= \frac{1}{2}$. What kind of figure is that? - Jun 17th 2009, 10:52 AMRachel.F
- Jun 17th 2009, 06:07 PMStroodle
$\displaystyle 2x^2+2y^2+x+y=0$

$\displaystyle 2(x^2+\frac{1}{2}x)+2(y^2+\frac{1}{2}y)=0$

$\displaystyle 2(x^2+\frac{1}{2}x+\frac{1}{16}-\frac{1}{16})+2(y^2+\frac{1}{2}y+\frac{1}{16}-\frac{1}{16})=0$

$\displaystyle 2[(x+\frac{1}{4})^2-\frac{1}{16}]+2[(y+\frac{1}{4})^2-\frac{1}{16}]=0$

$\displaystyle 2(x+\frac{1}{4})^2-\frac{1}{8}+2(y+\frac{1}{4})^2-\frac{1}{8}=0$

$\displaystyle 2(x+\frac{1}{4})^2+2(y+\frac{1}{4})^2=\frac{1}{4}$

$\displaystyle (x+\frac{1}{4})^2+(y+\frac{1}{4})^2=\frac{1}{8}$

$\displaystyle (x+\frac{1}{4})^2+(y+\frac{1}{4})^2=\left (\frac{1}{2\sqrt 2}\right )^2$

Which is a circle with it's centre at $\displaystyle \left ( -\frac {1}{4}, -\frac {1}{4}\right )$ with a radius of $\displaystyle \frac{1}{2\sqrt 2}$