Prove this without completing the square

[eqn](x_1+x_2)^2-4x_1x_2=(x_1-x_2)^2[/eqn]

The result follows trivially from this identity.

We all have our days, man.

im not the bloke who wrote that but:

[math]a^2 + 2ab + b^2 = (a + b)^2[/math]

based wolfram poster

The fact to the matter is that these equations are not numerically stable, please use different ones if you actually numerically calculate your results.
2/10, see me after class

show it then, if it's so trivial

[eqn](x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2[/eqn]
Using the identity, you can write both the sum and difference of the roots in terms of the coefficients. You just need to solve those two linear equations to get the individual roots. Trivial.

you are still completing the square here bro...

>without completing the square
>brainlet used completing the squared technique

AHHAHAHHHHHAHAHAHAH

This is not a logical statement and therefore cannot be proved. Try stating the assumptions and make it clear what you want to prove.

Protip: for the first equation to imply the second one, we need an assumption on the number a

Solve the system of equations

x1 * x2 = c/a;
x1 + x2 = -b/a

[math]\begin{align} x_1x_2 &= \frac{c}{a} \\ x_1+x_2 &= - \frac{b}{a} \\ x_1-x_2 &= \sqrt{(x_1+x_2)^2-4x_1x_2} = \sqrt{ \frac{b^2}{a^2}- 4 \frac{c}{a}} = \frac{1}{a} \sqrt{ b^2- 4ac } \\ x_1 &=\frac{1}{2}((x_1+x_2)+(x_1-x_2)) = \frac{1}{2}(- \frac{b}{a} + \frac{\sqrt{ b^2- 4ac }}{a} ) = \frac{-b+\sqrt{ b^2- 4ac }}{2a}\\ x_2 &=\frac{1}{2}((x_1+x_2)-(x_1-x_2)) = \frac{1}{2}(- \frac{b}{a} - \frac{\sqrt{ b^2- 4ac }}{a} ) = \frac{-b-\sqrt{ b^2- 4ac }}{2a} \end{align}[/math]

This

...

where does the result of "x1-x2=" come from

Let's recall from vieta that
[math] x_1 + x_2 = \frac{-b}{a}, x_1 x_2 = \frac{c}{a} [/math]

Lets square the first equation to get
[math] x_1^2 + 2x_1 x_2 + x_2^2 = \frac{b^2}{a^2} [/math]

And now let's substract four times the second equation to this to get:

[math] x_1^2 - 2x_1 x_2 + x_2^2 = \frac{b^2}{a^2} - \frac{4c}{a} \implies (x_1 - x_2)^2 = \frac{b^2}{a^2} - \frac{4c}{a} \implies x_1 - x_2 = \frac{\sqrt{b^2 - 4ac}}{a} [/math]

We now have an expression for [math] x_1 + x_2 [/math] and [math] x_1 - x_2 [/math] so now we can simply solve for them, for example
[math] x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} [/math]

>AHHAHAHHHHHAHAHAHAH
Mom let the hyaena use her computer again.

b^2=a
|b|=a^0.5
b=±(a^0.5)

First prove that [math]\mathbb{C}[/math] exists.

...

define what you mean by "exist" first

Show me a proof of existence with no hypotheses in any logic which isn't known to be inconsistent.

>It's the "ZFC is incosistent" autist again

>ZFC
Excuse me?

work in ZFC.
define N, construct Z, construct Q, construct R, construct C.
C exists, what's the problem?

Oops, forgot to mention "no outdated set-theoretic garbage".

there he goes again! haha, got me good!

work in ZFC.
define "proof", define "consistency", interpret ZFC in ZFC, demonstrate that a proof of "ZFC is consistent" exists.
ZFC is consistent, what's the problem?

Please provide a proof using either category theory (or a suitable type theory) or just first order logic. Sets need not apply.

>work in ZFC to talk about ZFC
genius
how about the usual proof using set theory that all mathematicians know and use? I mean, if you're not interested in math, then why the hell talk about it anyway?

>work in ZFC to talk about ZFC
Are you legitimately implying ZFC can't talk about ZFC?

not directly. if you want to get any information about ZFC from ZFC you need to work in a bigger theory.

>set theory
Outdated garbage which can hardly even be called mathematics at this point.
>you're not interested in math
My posts suggest otherwise.

your posts suggest your interest is in mathematical logic and applied shitposting
aka jokes, go do real math

>suggest your interest is in mathematical logic
Not really, it's just so that beings such as yourself can communicate with me. There aren't any other tools which you people know anyway.
>applied
So engineer-like garbage? Not interested.

>real math
Also, what are in your opinion some examples of it? Your perception of "real math" is quite warped if you think "set theory" has anything to do with it.

>engineer
I mean any serious math, you know? Analysis, Geometry, Algebra, whatever you want. But actual math.

I mean . Set theory is just some simple suitable foundation in which we can do the things we want to do. In the end, any other good foundation will have to fit all the math we have anyway.

>Analysis
That's more of a subfield of engineering.
>Geometry
>Algebra
So algebra? Yes, that's one of my interests as my posts heavily imply.

>Set theory
As in "ZFC" and the likes? Outdated garbage.

My main interest is in algebraic geometry and I know better than to call analysis a "subfield of engineering". Complex analysis in particular has deep connections to number theory and geometry.

Any foundations you give that are suitable to do algebra in will include (as an axiom or consequence) all of the axioms of ZFC.

>I know better than to call analysis a "subfield of engineering"
Apparently you don't. The type of "people" who study it are basically like engineers.
>Any foundations you give that are suitable to do algebra in will include (as an axiom or consequence) all of the axioms of ZFC.
Indeed, so might as well pick non-shit foundations.

honestly I hope you learn not to talk like that near any mathematicians or your career isn't going anywhere

either way you obviously agree that the axioms of ZFC are true in any good foundation, so you know how to build C

>honestly I hope you learn not to talk like that near any mathematicians
Most mathematicians from respectable fields already think this way, so it doesn't really need to be said out loud.
>axioms of ZFC are true
I never said anything about truth.

you know what "true" means in colloquial speech in mathematics, you're just being an ass at this point

To finish your proof, you'd need to know that a quadratic has at most two roots. To show that, just use the facts:

1) k is a root for a polynomial p(x) if and only if (x-k) divides p(x).

2) deg(p*q) = deg(p) + deg(q).

Together these force a polynomial of degree n to have at most n roots. And then since you've found two roots, you have them all.

wouldn't x-k result in 0 and then dividing that polynomial by 0 would be undefined?
no insults please

fpbp

By "a divides b" we mean "b = a*c for some c". So for example, in the integers 2 divides 6 since 6 = 2*3. For polynomials, an example would be (x-1) divides x^2 - 1 since x^2-1 = (x-1)*(x+1). So we don't actually divide and avoid problems like division by zero.

how can i construct the complex numbers algebraically?

R[x] / (x^2 + 1)

how can i construct R[x] algebraically?

take the usual graded algebra given by sequences in R, aka identify x by (0,1,0,0,...) and 1 by (1,0,0,0, ...)

what if I don't necessarily believe in these kinds of sequences? is there another algebraic construction?

You can probably do this by induction. First prove it for a=1, b=0, c=0. Then for positive a and nonnegative b, c prove that the given solution for (a, b, c) implies the solution for (a+1, b, c), then do the same for (a,b,c) => (a,b+1,c) and (a,b,c)=>(a,b,c+1). you can also do (a,b,c)=>(a,b-1,c) etc.

>You can probably do this by induction.
Won't work for real numbers.

then prove x1x2 = c/a and x1 + x2 = -b/a smartass

Sure, take all ordered n-tuplets with entries in R with the nth term nonzero and pretty much do the same thing.

This is one of the most entertaining arguments about numbers that I've ever witnessed. I didn't know nerds trying to out-nerd each other would be this funny.

expand the polynomial as a(x-x1)(x-x2) and match coefficients

see

take the vector space R^2 and induce an R-algebra structure by (1,0) identity and (0,1)^2 = (-1,0)

does it depend on R being a "field" or can i just treat it as a module over a ring? I do not necessarily believe in fields.
is there a construction which uses only the fact that R is a ring?