Prove that abcd = 2004

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Prime Newtons

Күн бұрын

Пікірлер: 61

@PrimeNewtons 7 ай бұрын

This is Vieta's Formula kzbin.info/www/bejne/j3qmlmBjea18a7csi=47ghMcgtTeylHNgQ

@florianbasier 7 ай бұрын

And if you don't know it, that's fine. You've just demonstrated that a,b,-c and -d are four roots of P(x)=x4-90x2+x+2004 so we know P(x)=(x-a)(x-b)(x+c)(x+d). If you now develop and associate the terms you have a system : 1=1 (terms in x4); c+d-a-b=0 (in x3); cd+ab-ac-ad-bc-bd=-90 (in x2); abc+abd-acd-bcd=1 (in x); abcd=2024 (in 1). You have answered the question and have enough data to compute a, b, c and d individually

@knupug 7 ай бұрын

I had no idea how to prove this. I only noticed that 45^2 is 2025 and when you subtract 21 you get 2004. And that if you look at abcd, you'd get something involving the square root of 45^4 and the square root of 21^2. How you get from that knowledge to the answer is beyond me.

@epikherolol8189 7 ай бұрын

Well because product of 4 roots is the constant term divided by leading coefficient.

@lcex1649 6 ай бұрын

Vista’s formula is just if you had (x-a)(x-b)(x-c)… where a,b,c,… are roots. When you multiply it out, there is a pattern and the second term is -1 (a+b+c+…), and the last term is (abc…).

@franolich3 6 ай бұрын

Need to prove that a, b, (-c), (-d) are distinct roots otherwise one cannot use Vieta's formula. Sure, the multiplicity of a root of a polynomial can be greater than 1 but it's just as possible in this case that 2 of these 4 numbers are equal and there is another root we are unaware of. So to prove the distinctness of these roots.... If a=0 then 0=sqrt(45-sqrt(21)) which is false so a≠0. Similarly b≠0, (-c)≠0, (-d)≠0. If a=b then: sqrt(45-sqrt(21-a)) = sqrt(45+sqrt(21-a)) => -sqrt(21-a) = sqrt(21-a) => 2.sqrt(21-a) = 0 => a=21 But putting this in the original equation give 21=sqrt(45) which is false, so a≠b. Similarly (-c)≠(-d). Since a and b are positive (defined as a square root and non-zero) and (-c) and (-d) are negative (for similar reasons) then a≠(-c), a≠(-d), b≠(-c), b≠(-d). So we have: a≠b. a≠(-c), a≠(-d), b≠(-c), b≠(-d) (-c)≠(-d) and this proves that the roots a, b, (-c), (-d) must be distinct.

@PrimeNewtons 6 ай бұрын

Thank you 👍

@zpf6288 4 ай бұрын

Vieta's formula is still valid for non-distinct roots, i.e. multiple roots with the same numerical value. The only thing one should not forget is to multiply them all and not just the numerically distinct roots. For example if a=4 and b=c=d=3, then one has abcd=4×3×3×3 and not just 4x3.

@Hussain-px3fc 4 ай бұрын

@@zpf6288exactly this is what I’m thinking but I see other people saying the same thing 😅

@Harrykesh630 7 ай бұрын

my first thought seeing this was, "NO I can't square these "

@0lympy 7 ай бұрын

8:50 Assuming that "if we have the same polynomial after double-squaring of all the 4 equations then each of its 4 roots is exactly a root of one of 4 source equations" looks not too obvious for me in all cases. What if we have different numbers and, thus, two zero roots (or different complex ones), for example? I feel that in common case it requires more accurate domain investigation for each source equation before doing these assumptions.

@razvandita2815 14 күн бұрын

this is kind of my concern as well. i am thinking that at least it should have been shown that a,b,c,d are all different

@Jon60987 7 ай бұрын

I am not familiar with the theorem you used. So IF you choose different roots for a,b,c, and d, then the product is 2004? I am guessing that a,b,c, and d are all different numbers, so they have to be different roots? A bit more explanation of this part would be appreciated.

@Uranyus36 7 ай бұрын

Essentially the polynomial we're dealing with here is P(x) = x^4 - 90x^2 + x + 2004. Since both a,b,c,d equates P(x) to 0, that means a,b,c,d are the four roots of P(x). Their values don't have to be all distinct from others, e.g. a can be equal to b or b = c = d. So long as they are both roots of P(x), we can apply Vieta's formula and find out their product is the constant term over the coefficient of the largest power term.

@0lympy 7 ай бұрын

There is a separate video about Vieta's theorem about a week ago there

@koktangri 6 ай бұрын

it's the Vieta's theorem, it's taught in middle school usually after you learn about the quadratic formula

@0lympy 6 ай бұрын

@@koktangri well, in the middle school it is usually limited to solve only quadratic polynomial, but the video is about the common case.

@oldjoec3710 2 ай бұрын

1. A little investigation with a spreadsheet shows that the value of a^4 - 90*a^2 - a + 2400 is STRICTLY POSITIVE for all real values of a (minimum value around +382). There ARE NO real roots. The correct answer is "No real solutions" to the abcd problem. 2.Even if we consider complex solutions, need to specify that a, b, c, d are unique. Nothing in the problem statement keeps me from saying that a, b, c, d are all the SAME root of the common equation. In that case, the abcd product would have to equal the fourth power of a, and a would have to be one of plus/minus 4th root of 2400, or i times plus/minus 4th root of 2400, none of which is an actual solution.

@pojuantsalo3475 7 ай бұрын

I tried to solve this by multiplying the square root expressions of a, b, c and d in order to get a massive square root expression where a, b, c and d are "mixed" and then studying what is needed to make the expression real, but doing that turned out so complicated I failed to solve this. Intuition is my strengths, but the solution you showed in this video is beyond my intuition I'm afraid even if I had tried more than 15 minutes to solve this.

@Reereez4795 7 ай бұрын

That’s the beauty of learning math. Now that you’ve seen the video, this technic can be a part of your problem-solving arsenal to use in the future! 😊

@gvssen 2 ай бұрын

As an extension we see that a+b+e+f = 0 or a+b - c -d = 0 which follows that a+b=c+d.

@KPunktFurry 7 ай бұрын

0:56 maby i will do but annyway an interesting equation! 2:59 ok till this point i would had done the same 5:00 logig it have to work becauce it is allways the same equantion! just with other variable names (except c, wait ok now it is more difficult) 5:57 ok i see the problem, now i am interested in the solution 7:17 using an other variable without chanching the equantion nice idear! 10:24 ok that is a bit difficult with out knowing this formular! but nice video again! and this times i subed ps: sorry for my bad english see you K.Furry

@aidan-ator7844 7 ай бұрын

Good question. I am not familiar with vieta's formula. Should the formula be fairly obvious or intuitive? If one hasn't been educated on certain theorems then are they supposed to figure them out by themselves?

@PrimeNewtons 7 ай бұрын

I explained it in a recent video with that title

@0lympy 7 ай бұрын

@@PrimeNewtons I'd suggest you adding a link (youtube has (i) in the corner of this video) to the related Vieta's video.

@PrimeNewtons 7 ай бұрын

@@0lympy thanks. I I just did

@15121960100 6 ай бұрын

there seems to be some issue in the logic of the solution

@HarmeetSingh-ou1eh 5 ай бұрын

Yes there is an issue

@ghtoostvilliers 6 ай бұрын

I thought of it in a way that original equation with 4 roots maybe can he written as in terms of variable x like :x=(45+ or -(21+ or -x)½)½ Thus x²=45+ or - (21+ or -x)½ (x²-45)²=21+ or -x x⁴-90x²+2025=21+ or - x x⁴-90x² + or - x +2004=0 Clearly product of all 4 roots taken at a time which are a,b,c,d is given as: abcd=2004 !💜

@domanicmarcus2176 6 ай бұрын

I don't understand. For "a" you got 2004, but for both "b" and "c" you are writing 2024. Is the problem supposed to be equal to 2004 or to 2024?

@ramennoodle9918 2 ай бұрын

i only see 2004 on the board

@marcgriselhubert3915 7 ай бұрын

The product of ALL the four complex solutions of the equation x^4 - 90.x^2 - x + 2024 = 0 is 2024, OK. But here is it certain (considering the means of definition of these numbers) that a, b, e and f are ALL the solutions of the equation ? If not, the result is false.

@epikherolol8189 7 ай бұрын

Yeah we must show that the polynomial contains no complex roots first

@iggberr 7 ай бұрын

I took exactly the same path to solve it. The only thing which is missing in your proof is the proof of the fact that that there are no equal values among abcd or abef. But it is easy. By that it is proven that there are no double roots.

@marcgriselhubert3915 7 ай бұрын

@@epikherolol8189 The equation has only 2 real solutions. So the given answer has two faults.

@Arkapravo 7 ай бұрын

Unique question!

@jimbeasley5319 7 ай бұрын

You didn’t show that a,b,c,d are the roots of the polynomial.

@alexlucky3338 6 ай бұрын

Exactly. I'm still confused. What I would highlight is that |Ai|=|Bi|=|Ci|=|Di| (meaning that moduli of the respective i-th roots of each equation are equal), but this does not provide for the product a·b·c·d to be equal to 2004. Can't see any clue yet. Would love to dig deeper into this 🙏🏻

@razvandita2815 14 күн бұрын

yes I have the same issue.

@dirklutz2818 6 ай бұрын

Tremendous!

@Menacé0nth3g0 6 ай бұрын

Nagtuturo na pala ng high math itong si apple de ap. Hahahhahahahhaha 🤣🤣🤣

@surendrakverma555 7 ай бұрын

Very good. Thanks 👍

@marcolima89 7 ай бұрын

Initially I thought the goal was to prove that a=2, b=0, c=0 and d=4 So that, abcd is 2004 😅

@KaiserBob99 7 ай бұрын

That's... not how this works

@aidan-ator7844 7 ай бұрын

This is real math, not goofy logic puzzles.😂

@marcolima89 7 ай бұрын

@@aidan-ator7844 i know it's real maths. I really don't get how what i said makes it a goofy puzzle. The hard part of this problem is in finding the roots of the equation. After that it can be the product of the roots, or the roots put together, or the sum of the roots, or anything else really. I think that's the fun of math, all problems or puzzles are valid, none of them is "goofy". Some problems don't have some practical application, but that doesn't matter.

@aidan-ator7844 7 ай бұрын

@@marcolima89 for me it has nothing to do with real-world practicality or not. It has to with the fact that it can only be the product of the roots. It cannot be the sum of the roots in any way. And in maths, we don't use letters when denoting the digits of a large number unless indicated earlier. An example would be some unknown digit represented by a letter in a summation.

@marcolima89 7 ай бұрын

@@aidan-ator7844 if the problem asked for a+b+c+d, you're telling me it couldn't be that? If the problem asked for the values of the individual roots its couldn't be that? I know that this specific problem didn't ask for any of that, but are you seriously telling me that if it asked for anything else it would have been considered a "goofy" logic puzzle? And logic puzzles are not goofy at all, they have value in developing skills also used in math.