# Thread: Will this equation ever be negative (proof)

1. There ya' are, MrF. That's what I was getting at, but I went to bed instead.

I was actualy going to rotate and eliminate the xy to try and see what we actually had, but...........

I like Moo and Oo 'completing the square' approaches. I would think there are many ways to show this is non-negative.

2. Very sorry. Don't know what was going on in my head. Basically, I wanted to get rid of 6xy by converting it (along with an $x^2$ and $y^2$ term) into the form $(a+b)^2$ from $a^{2} + 2ab + b^{2}$ where 2ab represents 6xy.

If 2ab = 6xy, possible values of a and b can be: $a = 2x; b = \frac{3}{2}y$ .

So we have: $4x^{2} + 6xy + \frac{9}{4}y^{2} = (2x + \frac{3}{2}y)^{2}$.

${\color{red}9x^{2}} + 6xy + {\color{blue}4y^{2}} - 3x - 2y + 1$
$= \underbrace{{\color{red}4x^{2} + 5x^{2}}}_{9x^{2}} + 6xy + \underbrace{{\color{blue}\frac{7}{4}y^{2} + \frac{9}{4}y^{2}}}_{4y^{2}} - 3x - 2y + 1$
$=\underbrace{{\color{red}4x^{2}} + 6xy + {\color{blue}\frac{9}{4}y^{2}}}_{\left(2x + \frac{3}{2}y\right)^{2} } + \underbrace{ {\color{red}5x^{2}} - 3x}_{\text{Factor}} + \underbrace{{\color{blue}\frac{9}{4}y^{2}} - 2y}_{\text{Factor}} + 1$

The rest follows.

3. Originally Posted by o_O
$=\underbrace{{\color{red}4x^{2}} + 6xy + {\color{blue}\frac{9}{4}y^{2}}}_{\left(2x + \frac{3}{2}y\right)^{2} } + \underbrace{ {\color{red}5x^{2}} - 3x}_{\text{Factor}} + \underbrace{{\color{blue}\frac{9}{4}y^{2}} - 2y}_{\text{Factor}} + 1$

The rest follows.
Ok but once you've factored, how can you state whether it is positive or not ?

Oh, by "factor" you mean complete the square ^^

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