Another one

[autumn topaz]

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[dreamy spire]

This has the form u + v - u^2 + uv - v^2 = 1, so:
u^2 - uv + v^2 - u - v + 1 = 0
Now, try bringing this to the canonical form.

[red briar]

i am unable to do that

[dreamy spire]

Well, start by finding a linear substitution of the for u = U + a, v = V + b that gets rid of linear terms.

[red briar]

i didnt understand

[dreamy spire]

Which part?

[red briar]

the linear substitution

[dreamy spire]

If you have a general second degree polynomial of two variables, there always exists a linear substitution like the one I wrote above that gets rid of at least one linear term.

Basically, we're trying to translate the respective quadratic curve so that it becomes centered at the origin.

[red briar]

oh ok

how do we apply it?

[dreamy spire]

Substitute u = U + a, v = V + b into the initial expression and group the terms. The coefficients of linear terms will be expressed in terms of a and b, so you can set up a system that they are both 0.

[red briar]

okay, i'll try

in like 12 hrs as i have to sleep and go to school

[tall trellis]

@red briar

Can you see that the equation is basically in the form of

$a^2+b^2+c^2-ab-bc-ca=0$

[stark bison]

dy/dx= ∏sin(jx).Σj/sin(jx)

[dreamy spire]

I don't think it has that form.

[tall trellis]

I think it has

[dreamy spire]

It does have the form u + v - u^2 + uv - v^2 = 1, though.

[tall trellis]

The equation can be rewritten like this

$(2^x)^2+(3^x)^2+1^2-3^x.1-2^x.3^x-2^x.1=0$$

[stark bison]

dy/dx= ∏sin(jx).Σj/sin(jx)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tall trellis]

now this is the form

[dreamy spire]

Ohh, interesting! I didn't look at it like that.

[tall trellis]

Now using the properties you will get x=0

[dreamy spire]

Well, we can bring that to the canonical form, too. Though, it's a bit harder than for two variables.

[tall trellis]

Yeah we can

[rancid torrent]

oh i see it. yeah if i multiply both sides by 2, i can get it in the form of:
(a - b)^2 + (b - c)^2 + (a - c)^2 = 0

and so a = b = c is the only real solution

[dreamy spire]

Oh, interesting! Not the canonical form, but a nice idea.

[red briar]

yeah so x=0

[dreamy spire]

Yup.

[tall trellis]

Yeah that equation is basically an identity as well

[tall trellis]

yeah

$a^2+b^2+c^2-ab-bc-ca=1/2[(a-b)^2+(b-c)^2+(c-a)^2]$