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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!

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    All caps is considered shouting on the internet - please don't do it again.2012-07-16
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    @ZevChonoles Sorry2012-07-16
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    Maybe on Facebook, sheesh!2015-10-24

1 Answers 1

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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?

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    Line two, where did the $2\dfrac{1}{2}$ originate from? And also, what happened to the $y$?!2012-07-16
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    @AustinBroussard I'm just dealing with $3x^2+3x$, you put the things together, =). The $2 \frac 1 2 $ is just to emphasize the middle term in $(a+b)^2 =a^2+\color{red}{2} b a +b^2$ which is in this case $1/2$ and $x$.2012-07-16
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    Henceforth $b=\dfrac{1}{2}$ ?? and if we're talking about $a^2+2ba+b^2$ , how does that correlate to $3x^2+3x=3(x^2+x)$2012-07-16
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    @AustinBroussard Try to think and re read what I wrote above and see if you can work it out yourself.2012-07-16
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    $2\dfrac{1}{2}$ comes from $\left(\dfrac{3}{2}\right)^2$ ? Isn't that how you get the middle term? Divide by $2$ and square it? But $\left(\dfrac{3}{2}\right)^2=2\dfrac{1}{4}$2012-07-16
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    And why is $\dfrac{1}{2}^2=\dfrac{3}{4}$ shouldn't it be $\dfrac{1}{4}$ ?2012-07-16
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    @AustinBroussard Have you recognized the $3$ in front of the brackets?2012-07-16
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    @draks but, why would the $3$ only be distributed into the $\dfrac{1}{4}$ and not the $3\left(x+\dfrac{1}{2}\right)^2$ ? oh and can you answer my other question. The question above the one I just asked. Thanks, just trying to clear up some confusion because I need to know HOW to do the problem, not just the answer. Thanks for the cooperation.2012-07-16
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    it is: $\displaystyle \color{red}3\left[\left( x+\frac1 2 \right)^2 -\left( \frac1 2 \right) ^2\right]=\color{red}3\left( x+\frac1 2 \right)^2- \frac{\color{red}3}{4}$ ... ok you're welcome.2012-07-16
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    Oh, right. Haven't slept yet. Um, can you address my previous question please.2012-07-16
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    Which one? ${ }$2012-07-16