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Standard forms: $y-b=A(x-a)^2$ or $x-a=A(y-b)^2$
$3x^2+3x+2y=0$

I honestly do not know how to start this problem. I have tried a lot of things and obviously not the right one. Can someone explain to me the first step and nothing more and I will edit with my new discoveries. Thanks!

  • 0
    Maybe on Facebook, sheesh!2015-10-24

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Since you have a factor with $x^2$, this anticipates the form you will be aiming for is

$y-b=A(x-a)^2$

So let's look at your eqn.:

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

We need to produce a perfect square with $3x^2+3x$. So, we can do the old completing the square trick:

$3x^2+3x=3(x^2+x)$

$3x^2+3x=3(x^2+2\frac 1 2 x)$

$3x^2+3x=3\left[x^2+2\frac 1 2 x+\left( \frac1 2 \right)^2 -\left( \frac1 2 \right) ^2\right]$

$3x^2+3x=3\left[\left( x+\frac1 2 \right)^2 -\left( \frac1 2 \right) ^2\right]$ $3x^2+3x=3\left( x+\frac1 2 \right)^2- \frac3 4$

Can you move on?

  • 0
    Which one? ${ }$2012-07-16