# Math Help - Quadratic Forms, Reduction

Im looking for a method of reducing a quadratic Q(x,y,z)

The example I have is Q(x,y,z) = 4x^2 + 14y^2 + 5z^2 + 16xy - 8xz - 20yz

I know the answer is: Q(x,y,z) = 4(x + 2y - z)^2 - 2y^2 + z^2 - 4yz

= 4(x + 2y - z)^2 - 2(y + z) + 3z^2

any advice would be greatly appreciated.

2. This is really tricky, but I would suggest (working backwards) you start with

$Q(x,y,z) = 4(x + 2y - z)^2 - 2y^2 + z^2 - 4yz$

and expand it out until you reach this

$Q(x,y,z) = 4x^2 + 14y^2 + 5z^2 + 16xy - 8xz - 20yz$

this is the answer in reverse and might give you some further insight.

3. Thats what I tried to try and see a link or methodology, but its the getting from the first Q(x,y,z) to the second that the question really requires.
I just cant see the train of thought process behind the reduction.

4. Originally Posted by tomevans1664
Im looking for a method of reducing a quadratic Q(x,y,z)

The example I have is Q(x,y,z) = 4x^2 + 14y^2 + 5z^2 + 16xy - 8xz - 20yz

I know the answer is: Q(x,y,z) = 4(x + 2y - z)^2 - 2y^2 + z^2 - 4yz

= 4(x + 2y - z)^2 - 2(y + z) + 3z^2

any advice would be greatly appreciated.
I have very little knowledge on the subject, but after doing some internet research, I'm curious about your given answer. The online literature (such as this Wikipedia article) seems to indicate we need the following form:

$q(x,y) = ax^2 + bxy + cy^2$

Using letters that don't overlap with the given letters, this is

$q(r,s) = ar^2 + brs + cs^2$

Letting $r = x + 2y - z$ and $s = z$ we obtain $rs = xz + 2yz - z^2 \not = y + z$.

This is all right, I suppose?

Edit: I just noticed that

$- 2y^2 + z^2 - 4yz \not = - 2(y + z) + 3z^2$

Did you make a typo somewhere?

5. Just double checked the answer and the solutions and its right (this is a past paper so I'm assuming there are no typos. Although the method you used is for binary quadratic equations here this is a ternary quad. equ. so im not sure if I'm able to make the substitution like that.

6. This is the link for my couse notes http://www.maths.qmul.ac.uk/~omj/MTH6140/la26.pdf there is a section in there on reduction of quadratic forms followed by an example but i cant seem to make th link between the theory to the example, feel free to try and make some sense of it...

7. Originally Posted by tomevans1664
Just double checked the answer and the solutions and its right (this is a past paper so I'm assuming there are no typos. Although the method you used is for binary quadratic equations here this is a ternary quad. equ. so im not sure if I'm able to make the substitution like that.
Okay, the internet literature on ternary quadratic forms seems sparser.

Well, being slightly lazy, I copied and pasted in Mathematica rather than checking by hand. (Click image to see full size.)

Unfortunately this does not match the original polynomial, and it's a bit hard to reverse engineer a flawed example.

8. Okay, after a brief look at the notes, I see that we wanted

Q(x,y,z) = 4(x + 2y - z)^2 - 2(y + z)^2 + 3z^2

which does indeed equal the original polynomial. I'll look a bit deeper, but I might not be able to help you with your original problem, I'll let you know.

9. well another example i have is:

Q(x,y,z)=3x^2+10y^2-5z^2+12xy-6xz-4yz

Q(x,y,z)=3(x+2y-z)^2 -2y^2+8yz-8z^2= 3(x+2y-z)^2 -2(y-2z)^2

10. oh yeah, typo, whoops...

11. Well I wasn't able to follow the matrix proof in the notes (I couldn't easily tell what things were defined as, etc.), but I have a pretty simple way to get the answer for your original problem.

It's basically just completing the square three times, in a strategic manner.

First we "collect" all the x terms, so that we'll have a squared term with an x in it, and other terms that don't have any x's in them.

So we need to combine

4x^2 + 16xy - 8xz

into one square. We complete as follows

4x^2 + 16xy - 8xz

= 4(x^2 + 4xy - 2xz)

= 4(x^2 + (4y - 2z) x)

Remember that we need to take half of the coefficient in front of x.

$= 4\left(x+\frac{4y-2z}{2}\right)^2 - 4\left(\frac{4y-2z}{2}\right)^2$

$= 4\left(x+2y-z\right)^2-4\left(2y-z\right)^2$

And then do the same for the remaining terms, collecting the y terms together, and finally collecting the z terms. (In general. But in this case the remaining terms $-2 y^2 - 4 y z + z^2$ are "nicer" to work with than the general case, and we can immediately see/guess that (y+z)^2 will get us approximately where we want to go).