Algebra x^2, y^2, xy, etc

[fiery lotus]

1. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
2. Wait patiently for a helper to come along.
3. Once someone helps you, say thank you and close the thread with:

   +close
   

4. Feel free to nominate the person for helper of the week in #helper-nominations
5. Do not ping the mods, unless someone is breaking the rules.
6. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[errant eagle]

Maybe try a first square with x,y and scalar, and another with y and scalar.

[civic oxide]

Completeing the squares give

$(2x-4)^2+2(y-2)^2+4xy-1$

[ornate spire]

dy/dx= ∏sin(jx).Σj/sin(jx)

[errant eagle]

You still have xy outside the squares

[civic oxide]

Yeah

[errant eagle]

I expected something like a²(x,y)+b²(y)+c, minimizing in y and then in x.

[spiral bluff]

You could calculate the gradient to locate the critical points (and so the local extrema considering R^2 is an open set and the function here is class C1) but maybe that’s too much

[errant eagle]

It is a secondary or high school homework.

[spiral bluff]

Yes never mind then

My bad it’s overkill

[errant eagle]

Otherwise one could use Lagrange multipliers.

[rapid schooner]

"I tried completing the square, not sure if I did something wrong or not"
Show us how you completed the square

Complete the square with respect to x and treat y as if it was a constant ( you can also complete with respect to y by treating x as constant too, I just think it will be more messy), then complete the square with what is left

[rapid schooner]

You will get something that looks like this above ^

[errant eagle]

There would be 3 terms in a and 2 in b.

[rough dirge]

Start by bringing the expression to the canonical form. It's better to make such a substitution x = X + a, y = Y + b that the linear terms disappear, then it will be easier.

[wild gyro]

any of this help?

[errant eagle]

Thank you for your effort. I was rather thinking of something like a=px+qy+r and b=sy+t

[rough dirge]

Sorry, not quite sure what you're doing. My approach would either be what I described above or the following.
4x^2 + 4xy + 2y^2 - 16x - 4y + 23 = (2x + y - 4)^2 + y^2 + 4y + 7 = (2x + y - 4)^2 + (y + 2)^2 + 3
And now it's easy.

[wild gyro]

Is it possible you can attempt at solving it

I tried another approach

and got 3, -2

I need something to verify it with

[rough dirge]

Yup, that's correct.

[wild gyro]

Ok thx

[rough dirge]

Well, from the expression above it's quite obvious.

[wild gyro]

😭

lmaoo

im too slow for this math discord lmao

[rough dirge]

No worries!

Do you see how I was able to complete the squares?

[wild gyro]

Yes yes

[rough dirge]

Nice!

[wild gyro]

Thanks sm

[rough dirge]

You're welcome!

[wild gyro]

If its not too much to ask

I dont happen to have much luck on this either 😭

Any Idea how I can get around it?

[rough dirge]

Oh, I think I've seen this problem before.
Substitute w into the second equation, then multiply everything my xyz(x + y + z). Then, try grouping some terms. Or just look at the result reeeeally carefully.

Spoiler: it factors very nicely.

[wild gyro]

hmm

not sure i follow to well 😭

so essentially

what your saying is

yea im lost 😭

can we go through this step by step by chance?

[rough dirge]

What step is the problem? I don't really know how to say it in other words.

[wild gyro]

like

Im not sure how to evaluate it

[rough dirge]

What expression did you get after you substituted w and multiplied by xyz(x + y + z)?

[wild gyro]

w(x+y+z)

wait no mb

[rough dirge]

What? There won't be w after you do that.

[wild gyro]

xyz(w)

[rough dirge]

Uh... No.

[wild gyro]

hold on 😭 im so confused

[rough dirge]

When we substitute w into the second equation, we get 1/x + 1/y + 1/z = 1/(x + y + z). Now, multiply everything by xyz(x + y + z).

[wild gyro]

Ohh

I see what you mean now

my bad I was mistaken completely

hold on

[wild gyro]

where do i move from this point?

[rough dirge]

What did you get?

[wild gyro]

how did you happen to move from 1/x + 1/y + 1/z = 1/(x+y+z) to xyz(x+y+z)?

[rough dirge]

Well, xyz(x + y + z) is the common denominator. So, we multiply by it.

[wild gyro]

so

would i be bringing myself to a point like

yz/xyz +xz/xyz + xz/xyz = 1/x+y+z

then multiplying out x+y+z

to make it

[rough dirge]

Well, you also need to multiply both parts by xyz, too.

[wild gyro]

right

[rough dirge]

That way the whole thing becomes a polynomial.

[wild gyro]

xyz/x+y+z = xyz*x+y+z(yz + xy + xz)

?

[rough dirge]

Wait, how did this become a double equality?

[wild gyro]

sorry no

my bad

lmao

this is the step im at right now

wait no

[rough dirge]

We have:
1/x + 1/y + 1/z = 1/(x + y + z)
Multiplying by xyz(x + y + z) gives:
(x + y + z)(xy + xz + yz) = xyz
Now, expand the brackets.

[wild gyro]

xyz = xyz*x+y+z(yz + xy + xz)

[wild gyro]

Oh right

Because we cancel out xyz when multiplying it on top

my bad

from expanding the brackets

im here

[rough dirge]

No, that's incorrect. There should be way more terms.

[wild gyro]

oh shit

wait

how so

[rough dirge]

Also, your expression isn't symmetric in x, y, z, while it clearly should be.

[wild gyro]

x+y+z(xy+xz+yz)=xyz

expanding the brackets

[rough dirge]

(x + y + z)(xy + xz + yz), not x + y + z(xy + xz + yz).

[wild gyro]

oh okok

right

but

arent i just doing

x times xy, x times xz, x times yz

[rough dirge]

Uh, no...

[wild gyro]

what am i meant to do?

[rough dirge]

You have three terms in the first brackets and three in the second, so there should be nine terms in total (before grouping).

[wild gyro]

yeah

[wild gyro]

this is 9 terms is it not?

[rough dirge]

Ohh.

Wait, sorry.

You just didn't write the plus on each new row, so I thought those were separate expressions.

[wild gyro]

Ohhh LOL

Npnp

thats my bad tbh

I shouldve explained that

[rough dirge]

No worries! All good.

So, now you can see that you have xyz in several places, so you can simplify a bit.

[wild gyro]

right

brings me to 7 terms

= xyz btw

[rough dirge]

Well, you can bring it to the left, which gives (...) + 2xyz = 0.

Now, here's the trick.

[wild gyro]

heres where im at rn

[rough dirge]

Split the terms into two groups:
(x^2 y + x^2 z + xz^2 + xyz) + (y^2 z + yz^2 + xy^2 + xyz) = 0

Then factor x from the first term and y from the second. What do you see?

[wild gyro]

I see yes

essentially quadratics correct?

[rough dirge]

Well, yes, but that isn't the whole point.

What do you get when you factor x and y like that?

[wild gyro]

2 individual terms...

im not sure truthfully

i notice that if you factor x out

and y from the second

that you get

one term just inclusive of y and z

and the other inclusive of

x and z

[rough dirge]

What I mean is that you get the same expression:
x(xy + xz + z^2 + yz) + y(xy + xz + z^2 + yz) = 0

[wild gyro]

right

[rough dirge]

So, we get:
(x + y)(xy + xz + z^2 + yz) = 0

Now, try working with xy + xz + z^2 + yz.

[wild gyro]

x(y+z) + z(y+z)

[rough dirge]

Right, so?

[wild gyro]

(x+z) (y+z)

[rough dirge]

Nice! So, what does the whole thing become?

[wild gyro]

(x+z)(y+z)(x+y)

[rough dirge]

Most importantly, that equals 0.

[wild gyro]

right

[rough dirge]

So, there you go. That gives you conditions on variables.

I'll leave the combinatorics to you, as I kinda suck at it.

Remember that due to the initial second equation, x, y, z and x + y + z can't be 0.

[wild gyro]

yes thats true

[wild gyro]

do u happen to know the final?

[rough dirge]

No, sorry. In the previous discussion I was mainly interested in the factorization.

[wild gyro]

+close