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/sci/ - Science & Math

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9285817 No.9285817 [Reply] [Original]

prove this without completing the square

>> No.9285821

Substitute the solution as an anzats and show it satisfies the quadratic.

>> No.9285855

>>9285817
http://www.wolframalpha.com/input/?i=a((-b+%2B+sqrt(b%5E2+-+4*a*c))%2F(2a))%5E2+%2B+b((-b+%2B+sqrt(b%5E2+-+4*a*c))%2F(2a))+%2B+c

http://www.wolframalpha.com/input/?i=a((-b+-+sqrt(b%5E2+-+4*a*c))%2F(2a))%5E2+%2B+b((-b+-+sqrt(b%5E2+-+4*a*c))%2F(2a))+%2B+c

>> No.9286477

>>9285817

The product of the roots is c/a, and their sum is -b/a. Hence the square of their sum is b^2/a^2, and their difference is plus or minus [math]\sqrt{(b^2 - 4ac)}/a^2[/math]. QED

>> No.9286569
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9286569

>>9285855
>http://www.wolframalpha.com/input/?i=a((-b+-+sqrt(b%5E2+-+4*a*c))%2F(2a))%5E2+%2B+b((-b+-+sqrt(b%5E2+-+4*a*c))%2F(2a))+%2B+c

>> No.9286790
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9286790

>>9285817

Tchirnhausen transformation applies to the low-degree solvable cases.

>> No.9286803

>>9285817
[math]
\begin{align*}
ax^2 + bx +c &= 0 \\
ax^2 + bx &= -c \; \; \; \; | \; \cdot 4a \\
4a^2x^2 + 4abx &= -4ac \; \; \; \; | \; +b^2 \\
4a^2x^2 + 4abx +b^2 &= b^2 -4ac \\
(2ax + b)^2 &= b^2 -4ac \\
2ax + b &= \pm \sqrt{b^2 - 4ac} \\
2ax &= -b \pm \sqrt{b^2 - 4ac} \\
\displaystyle
x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\end{align*}
[/math]

>> No.9286809

>>9286803
How did you go from the 4th to 5th step?

>> No.9286824

>>9286809
Not him, but it is basic factorization...

>> No.9286842

>>9286824
I got it. Apologies for degeneracy.

>> No.9286856

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

The result follows trivially from this identity.

>> No.9286867
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9286867

>>9286842
We all have our days, man.

>> No.9286869

>>9286809
im not the bloke who wrote that but:

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

>> No.9286879

>>9285855
based wolfram poster

>> No.9286908

>>9285817
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

>> No.9286930

>>9286856
show it then, if it's so trivial

>> No.9286982

>>9286930
[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.

>> No.9287040

>>9286803
you are still completing the square here bro...

>> No.9287629

>>9286803
>without completing the square
>brainlet used completing the squared technique

>> No.9288725

>>9286982
AHHAHAHHHHHAHAHAHAH

>> No.9288734

>>9285817
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

>> No.9289138

Solve the system of equations

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

>> No.9289175

>>9288725
>>9289138
[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]

>> No.9289729

>>9289175
>>9289138
This

>> No.9290924
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9290924

>>9286477

>> No.9290966

>>9289175
where does the result of "x1-x2=" come from

>> No.9290969

>>9285817
>>9285817
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]

>> No.9292143

>>9288725
>AHHAHAHHHHHAHAHAHAH
Mom let the hyaena use her computer again.

>> No.9292154

>>9286809
b^2=a
|b|=a^0.5
b=±(a^0.5)

>> No.9292186

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

>> No.9292219

>>9290966
>>9286856

>> No.9292220

>>9292186
define what you mean by "exist" first

>> No.9292370

>>9292220
Show me a proof of existence with no hypotheses in any logic which isn't known to be inconsistent.

>> No.9292405

>>9292370
>It's the "ZFC is incosistent" autist again

>> No.9292420
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9292420

>>9292405
>ZFC
Excuse me?

>> No.9292434

>>9292420
work in ZFC.
define N, construct Z, construct Q, construct R, construct C.
C exists, what's the problem?

>> No.9292447

>>9292434
Oops, forgot to mention "no outdated set-theoretic garbage".

>> No.9292451

>>9292447
there he goes again! haha, got me good!

>> No.9292452

>>9292434
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?

>> No.9292459

>>9292451
Please provide a proof using either category theory (or a suitable type theory) or just first order logic. Sets need not apply.

>> No.9292466

>>9292452
>work in ZFC to talk about ZFC
genius
>>9292459
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?

>> No.9292469

>>9292466
>work in ZFC to talk about ZFC
Are you legitimately implying ZFC can't talk about ZFC?

>> No.9292474

>>9292469
not directly. if you want to get any information about ZFC from ZFC you need to work in a bigger theory.

>> No.9292481
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9292481

>>9292466
>set theory
Outdated garbage which can hardly even be called mathematics at this point.
>you're not interested in math
My posts suggest otherwise.

>> No.9292485

>>9292481
your posts suggest your interest is in mathematical logic and applied shitposting
aka jokes, go do real math

>> No.9292493

>>9292485
>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.

>> No.9292495

>>9292485
>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.

>> No.9292496

>>9292493
>engineer
I mean any serious math, you know? Analysis, Geometry, Algebra, whatever you want. But actual math.

>> No.9292503

>>9292495
I mean >>9292495. 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.

>> No.9292519

>>9292496
>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.

>>9292503
>Set theory
As in "ZFC" and the likes? Outdated garbage.

>> No.9292532

>>9292519
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.

>> No.9292557

>>9292532
>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.

>> No.9292563

>>9292557
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

>> No.9292568

>>9292563
>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.

>> No.9292581

>>9292568
you know what "true" means in colloquial speech in mathematics, you're just being an ass at this point

>> No.9292631

>>9285821
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.

>> No.9292638

>>9292631
wouldn't x-k result in 0 and then dividing that polynomial by 0 would be undefined?
no insults please

>> No.9292641

>>9285821
fpbp

>> No.9292653

>>9292638
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.

>> No.9292675

how can i construct the complex numbers algebraically?

>> No.9292682

>>9292675
R[x] / (x^2 + 1)

>> No.9292691

>>9292682
how can i construct R[x] algebraically?

>> No.9292703

>>9292691
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, ...)

>> No.9292756

>>9292703
what if I don't necessarily believe in these kinds of sequences? is there another algebraic construction?

>> No.9292799

>>9285817
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.

>> No.9292906

>>9292799
>You can probably do this by induction.
Won't work for real numbers.

>> No.9292989

>>9289175
then prove x1x2 = c/a and x1 + x2 = -b/a smartass

>> No.9292997

>>9292756

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

>> No.9293017

>>9292989

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.

>> No.9293053

>>9292989
expand the polynomial as a(x-x1)(x-x2) and match coefficients

>> No.9293056

>>9292989

see >>9286982

>> No.9293065

>>9292756
take the vector space R^2 and induce an R-algebra structure by (1,0) identity and (0,1)^2 = (-1,0)

>> No.9293092

>>9293065
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?