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

1. ## Will this equation ever be negative (proof)

Hello,

I have a discrete math assignment which says:

Is the following statement true or false? Explain.
For any real numbers x, y, the number 9x2 + 4y2 + 6xy − 3x − 2y + 1 is not negative.

I am pretty sure that this is true (I wrote a program that went through a lot of the possibilities and it was never negative). However, I do not have a vigorous proof that this is true. If anyone can help me get started on this proof it would be great. I talked to my prof and he said the best way was to not break it into cases, and that the other way is really easy. I apparently just can't see it. Thanks in advance for anyones help into this. I do not necessarily want the proof done for me, but even an idea where to start would be great.

2. Hello,

Originally Posted by SaxyTimmy
Hello,

I have a discrete math assignment which says:

Is the following statement true or false? Explain.
For any real numbers x, y, the number 9x2 + 4y2 + 6xy − 3x − 2y + 1 is not negative.

I am pretty sure that this is true (I wrote a program that went through a lot of the possibilities and it was never negative). However, I do not have a vigorous proof that this is true. If anyone can help me get started on this proof it would be great. I talked to my prof and he said the best way was to not break it into cases, and that the other way is really easy. I apparently just can't see it. Thanks in advance for anyones help into this. I do not necessarily want the proof done for me, but even an idea where to start would be great.
Are you familiar with Gauss ?

Nevermind, we'll just apply high school techniques then

$9x^2+4y^2+6xy-3x-2y+1$

Let's rearrange it :

${\color{red}(3x)^2+x(6y-3)}+4y^2-2y+1$

Red thing (we're going to complete the square) :

\begin{aligned} {\color{red}=} (3x)^2+2 \cdot (3x) \cdot (y-\frac 12) \\
{\color{red}=} \left(3x+y-\frac 12\right)^2-\left(y-\frac 12\right)^2 \\
{\color{red}=}\left(3x+y-\frac 12\right)^2-(y^2-y+\frac 14) \end{aligned}

The original expression is now :

${\color{red}\left(3x+y-\frac 12\right)^2}-y^2+y-\frac 14+4y^2-2y+1$

$={\color{red}\left(3x+y-\frac 12\right)^2}+{\color{blue}3y^2-y+\frac 34}$

3. $9x^{2} + 4y^{2} + 6xy - 3x -2y + 1$
$= 9x^{2} + 6xy + 4y^{2} - (3x + 2y) + 1$ (regrouped a bit)
$= {\color{red}(3x)^{2} + 6xy + (2y)^{2}} - (3x + 2y) + 1$

For the red, recall that $(a + b)^{2} = a^{2} + 2ab + b^{2}$. So factoring the red gives us
$= (3x + 2y)^{2} - (3x + 2y) + 1$

Now if you know anything about parabolas, then let $a = 3x + 2y$

Then we have $y = a^{2} - a + 1$ which is a parabola that never crosses the x-axis, something you can easily prove.

4. Originally Posted by o_O
$9x^{2} + 4y^{2} + 6xy - 3x -2y + 1$
$= 9x^{2} + 6xy + 4y^{2} - (3x + 2y) + 1$ (regrouped a bit)
$= {\color{red}(3x)^{2} + 6xy + (2y)^{2}} - (3x + 2y) + 1$

For the red, recall that $(a + b)^{2} = a^{2} + 2ab + b^{2}$. So factoring the red gives us
$= (3x + 2y)^{2} - (3x + 2y) + 1$

Now if you know anything about parabolas, then let $a = 3x + 2y$

Then we have $y = a^{2} - a + 1$ which is a parabola that never crosses the x-axis, something you can easily prove.

I thought the same thing, however, when I expanded
$(3x + 2y)^{2} - (3x + 2y) + 1$

I get
$9x^{2} + 4y^{2} + 12xy -3x -2y+1$

and that is not the original equation (its 6xy, not 12xy in the original equation). Which is because "2ab" would be $2*3x*2y$ which is equal to 12xy.

PS. I really do appreciate your help though.

5. What do you think of mine ?

I continue :

$
={\color{red}\left(3x+y-\frac 12\right)^2}+{\color{blue}3y^2-y+\frac 34}
$

$=\left(3x+y-\frac 12\right)^2+3y^2{-\color{magenta}3y}+\frac 34{+\color{magenta}2y}$

$=\left(3x+y-\frac 12\right)^2+3(y^2-y+\frac 14)+2y$

$=\left(3x+y-\frac 12\right)^2+3\left(y-\frac 12\right)^2+2y$

So my final thought is that there is a typo or a mistake because of 2y...

We can use the same argument, just with a little modification:
${\color{red}9x^{2}} + 6xy + {\color{blue}4y^{2}} - 3x - 2y + 1$
$= {\color{red}4x^{2} + 5x^{2}} + 6xy + {\color{blue}y^{2} + 3y^{2}} - 3x - 2y + 1$
$= \underbrace{{\color{red}4x^{2}} + 6xy + {\color{blue}y^{2}}}_{\text{Factor}} + \underbrace{{\color{red}5x^{2}} - 3x}_{\text{Factor}} + \underbrace{{\color{blue}3y^{2}} - 2y}_{\text{Factor}} + 1$

For the latter 2, you'll have to complete the square.

7. I am still confused about yours.... my weakest point in math is completing the square (I know I know... its the easiest thing to do... but its the one thing I can not do). However, I think you might have made a mistake in your square completion on the first line. I am still trying to comprehend it.. but the "dreaded" square completion is taking me time to try to figure out :P

Edit: I guess I mean between the first and second line.
Edit: There really is no error... like I said Completing Square + me = weak

8. I'll rewrite all this stuff (and change my second post)

Originally Posted by Moo
$9x^2+4y^2+6xy-3x-2y+1$

Let's rearrange it :

${\color{red}(3x)^2+x(6y-3)}+4y^2-2y+1$

Red thing (we're going to complete the square) :

\begin{aligned} &{\color{red}=} (3x)^2+2 \cdot (3x) \cdot (y-\frac 12) \\
&{\color{red}=} (3x)^2+2 \cdot (3x) \cdot (y-\frac 12)+(y-\frac 12)^2-(y-\frac 12)^2 \ \text{added this line} \\
&{\color{red}=} \left(3x+y-\frac 12\right)^2-\left(y-\frac 12\right)^2 \\
&{\color{red}=} \left(3x+y-\frac 12\right)^2-(y^2-y+\frac 14) \end{aligned}

The original expression is now :

${\color{red}\left(3x+y-\frac 12\right)^2}-y^2+y-\frac 14+4y^2-2y+1$

$={\color{red}\left(3x+y-\frac 12\right)^2}+{\color{blue}3y^2-y+\frac 34}$
Originally Posted by Moo
What do you think of mine ?

I continue :

$
={\color{red}\left(3x+y-\frac 12\right)^2}+{\color{blue}3y^2-y+\frac 34}
$

$=\left(3x+y-\frac 12\right)^2+(\sqrt{3}\cdot y)^2-2 \cdot (\sqrt{3}\cdot y)\cdot \frac{1}{2 \sqrt{3}} +\frac 34$

$=\left(3x+y-\frac 12\right)^2+(\sqrt{3}\cdot y)^2-2 \cdot (\sqrt{3}\cdot y)\cdot \frac{1}{2 \sqrt{3}}+(\frac{1}{2\sqrt{3}})^2- (\frac{1}{2\sqrt{3}})^2+\frac 34$

$=\left(3x+y-\frac 12\right)^2+\left(\sqrt{3}\cdot y-\frac{1}{2 \sqrt{3}} \right)^2-\frac{1}{12}+\frac 34$

$=\left(3x+y-\frac 12\right)^2+\left(\sqrt{3}\cdot y-\frac{1}{2 \sqrt{3}} \right)^2+\frac{8}{12}$
All terms are positive;

I am dumb
Nope.

9. Thanks a bunch moo... you are obviously a genius ^_^. As per your post o_O I am still trying to figure it out... which I guess it means its pretty genius too. You guys are awesome.

(PS. Also, thanks for saying I am not dumb lol)

10. Originally Posted by o_O

We can use the same argument, just with a little modification:
${\color{red}9x^{2}} + 6xy + {\color{blue}4y^{2}} - 3x - 2y + 1$
$= {\color{red}4x^{2} + 5x^{2}} + 6xy + {\color{blue}y^{2} + 3y^{2}} - 3x - 2y + 1$
$= \underbrace{{\color{red}4x^{2}} + 6xy + {\color{blue}y^{2}}}_{\text{Factor}} + \underbrace{{\color{red}5x^{2}} - 3x}_{\text{Factor}} + \underbrace{{\color{blue}3y^{2}} - 2y}_{\text{Factor}} + 1$

For the latter 2, you'll have to complete the square.
hey Moo,

the only thing I see that I don't like about this is that I don't think that there is a way to factor $4x^{2} + 6xy + y^{2}$ nicely.

I really do thank you both for your help gentlemen.

11. Originally Posted by SaxyTimmy
hey Moo,

the only thing I see that I don't like about this is that I don't think that there is a way to factor $4x^{2} + 6xy + y^{2}$ nicely.
Actually, that's why I proposed you to complete the squares ^^

The technique is to isolate the terms containing x. Then completing the square.
You'll be left only with terms containing y and constants... So complete again the square.

It's called "Gauss reduction", litteral translation from French... I don't know the name in English

(PS. Also, thanks for saying I am not dumb lol)
Well, no one is dumb. All this working comes with habits

I really do thank you both for your help gentlemen.

12. Ah, thanks. That makes sense. I think it is actually called Gauss reduction in English too (I have heard of it... but was not 100% sure what it was). I will have to try that to see what comes of it. When I started doing it I just started from the left and was like hmm... this doesn't work.

Also, thanks gentlewoman as well. I am really sorry... I usually do use both... I am ashamed

13. Originally Posted by SaxyTimmy
hey Moo,

the only thing I see that I don't like about this is that I don't think that there is a way to factor $4x^{2} + 6xy + y^{2}$ nicely.

I really do thank you both for your help gentlemen.
You cannot factor it nicely

Rearanging we see that can be

$4x^2+4xy+y^2+2xy\Rightarrow{(2x+y)^2+2xy}$

14. If you take note, this is some sort of conic section. The discriminant is less than 0 so it must be an ellipse, circle, a point, or has no graph.

$cot2{\theta}=\frac{A-C}{B}=\frac{5}{6}$

If we rotate to shed the xy term, it appears we have an ellipse. I didn't delve into it too deep though.

15. Originally Posted by galactus
If you take note, this is some sort of conic section. The discriminant is less than 0 so it must be an ellipse, circle, a point, or has no graph.

[snip]
Since I'm no good at algebra, that was my first thought as well But I think it needs the following follow-through. Briefly:

Consider $9x^2 + 4y^2 + 6xy - 3x - 2y + 1 = -k$, where k > 0.

To find the curve (if any) that this equation defines (that is, set of points (if any) that satisfy the equation):

a = 9, b = 3, c = 4 so the minor discriminant $b^2 - ac < 0$ therefore the curve is either an ellipse, a single point or there's no curve.

No curve if (and I can't be bothered doing the latex) the product of the determinant of the matrix

9, 3, -1.5
3, 4, -1
-1.5, -1, 1 + k

(that is, the major discriminant) and a + c is greater than zero:

(13) (27(k + 2/3)) is greater than zero when k > -2/3. Since by definition k > 0, the product is always greater than zero.

Therefore $9x^2 + 4y^2 + 6xy - 3x - 2y + 1$ cannot be less than zero.

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