Math Olympiad | A Nice Algebra Problem | How to solve for 'X' and 'Y' in this problem?

Рет қаралды 80,262

Күн бұрын

Пікірлер: 43

@johnmccullagh2705 10 ай бұрын

Good solid algebra. Yes a bit slow, but for someone like myself who has not done much serious math since student days (half a century ago now), better to include all the steps rather than skip some. And yes, the obvious real solutions from simple inspection are x=2, y=1 and x=1, y=2, however that's not the point of the exercise. Whether it's a good Math Olympiad problem or not, I'm afraid I can't judge.

@TWJRPGGamming Жыл бұрын

x=2.y=1 x=1,y=2

@neel333neel 9 ай бұрын

thanks

@learncommunolizer 9 ай бұрын

You're welcome!

@KipIngram 10 ай бұрын

By "looking at it and thinking for a second," x=2, y=1 and x=1, y=2 are solutions.

@АскидаСалимова 9 ай бұрын

Это же устно решается, сразу в глаза бросается, что х=1, у=2 или х=2, у=1.

@Misha-g3b 9 ай бұрын

Hо система может иметь др. решения.

@catiaman208 Жыл бұрын

this has sipleset way since the numbers of X AND Ymyst be integer X+Y=3 RESULT X=1 and Y=2 X5+y5=1+2^5=1+32=33

@ryanchiang9587 Жыл бұрын

x = 2, y = 1 or vice cersa

@guyhoghton399 Жыл бұрын

You can use the identity _uₘ₊ₙ = uₘuₙ - uₘ₋ₙ(xy)ⁿ_ where _uₙ = xⁿ + yⁿ_ Thus: _u₀ = 2, u₁ = 3, u₅ = 33_ _u₂ = u₁² - u₀(xy)¹ = 9 - 2xy_ _u₃ = u₂u₁ - u₁(xy)¹ = 3(9 - 2xy) - 3xy = 27 - 9xy_ _u₅ = 33 = u₃u₂ - u₁(xy)² = (9 - 2xy)(27 - 9xy) - 3(xy)²_ ∴ _243 - 135xy + 15(xy)² = 33_ ⇒ _(xy)² - 9xy + 14 = 0_ ⇒ _(xy - 2)(xy - 7) = 0_ ∴ _xy = 2_ or _7_ Note that _x_ and _y_ are the roots of _t² - (x + y)t + xy = 0_ Case (i): _xy = 2_ _t² - 3t + 2 = 0_ ⇒ _(t - 1)(t - 2) = 0_ ∴ _x, y = 1, 2_ in either order. Case (ii): _xy = 7_ _t² - 3t + 7 = 0_ ⇒ _t = ½[3 ± √(9 - 28)]_ ∴ _x, y = ½(3 + √19), ½(3 - √19)_ in either order.

@Verdiw Жыл бұрын

I graph it in demos , only 2 real sol Edit : Sadly at the last part you messed up 9-28 is -19 not 19 so 2 complex roots

@neilmccafferty5886 Жыл бұрын

you can see it must be 2 and 1 immediately.

@АндрейКожевников-о8й Жыл бұрын

Комплексные корни не интересны.

@neilmccafferty5886 Жыл бұрын

@@АндрейКожевников-о8й has to be integers, both greater than 1. simple.

@Verdiw Жыл бұрын

@@neilmccafferty5886Doesn’t need to be integer

@rafielmesaias 11 ай бұрын

We have an equation of grade 5. But you have just four roots or solutions por x and y

@GFlCh 10 ай бұрын

While it might be nice to see how to work out another solution, it's possible any additional solution would be extraneous, or a duplicate of an existing solution. But most importantly, any valid solution for the problem as a whole, has to work in both equations, and another solution to the "x⁵ + y⁵=33" equation, if it exists, might not work in the "x + y = 3" equation.

@Misha-g3b 9 ай бұрын

Such system has 5 solutions or less.

@prayitsmg7086 Жыл бұрын

X = 1 , Y = 2 or X = 2 , Y = 1

@markterribile6948 Жыл бұрын

Who is your audience? Your audience is people who can appreciate and learn what you are showing them. Such people must have mastered high school algebra, and have presumably mastered grade school arithmetic, preparing them to learn the more advanced techniques and principles you are showing them. They--we--don't need to see basic numerical division belabored. They--we--don't need the detailed steps of factoring a quadratic. With the factors given, we need about three seconds to verify them, then we can move on. The value in these solutions is the overall scheme, recognizing the larger patterns that we can use to pry the problem apart. But you bury those steps in the minutia that we have already mastered, making it hard to find the precious grains of gold mixed with the dross of first year algebra and fourth grade arithmetic. These are brilliant solutions. The presentations would be so much better if you demonstrated the brilliance and moved routinely through the routine mechanics.

@heniwatisetiono6995 Жыл бұрын

x=1, y=2 or x=2, y=1

@sudarshandutta2498 7 ай бұрын

X= 2/1, Y=2/1.

@wongkienloon 11 ай бұрын

33+3+33+3=72 66+6=72 (66/72) + (6/72)=1=100%

@ArashHormozi-xx3mu Жыл бұрын

X:2 y:1 or x:1 y:2

@is7728 Жыл бұрын

Can anyone explain why there are only 4 roots, not 5?

@Verdiw Жыл бұрын

An degree n equation doesn’t need to have n distinct root , some can be the same so the amount of root is not more than n

@is7728 Жыл бұрын

@@Verdiw I know that so which one is the double root?🌲

@nemesiochupaca5024 Жыл бұрын

Porque si sustituyes y=3-x Te va a quedar x^5+(3-x)^5-33=0 y el término quíntico se irá, quedando un polinomio de grado 4

@arturmadatyan4892 Жыл бұрын

Очень простое уровнение, почему так мучается?

@Норайр-у7т 10 ай бұрын

Надо немножко подумать,потом мучатся,точно !!!

@Bertin-q3y 8 ай бұрын

(1;2) (2;1)

@Misha-g3b 9 ай бұрын

(1, 2), (2, 1). They €R².

@Bertin-q3y Жыл бұрын

X=2 et Y=1

@BonchoIvanov-b1u 11 ай бұрын

Supposed to be 5 roots

@CrYou575 10 ай бұрын

If you set x = 3-y using equation 2 then the highest term of x^5 is - y^5, which cancels out with y^5 in equation 1. This reduces the problem to solving a quartic and results in four roots.

@Misha-g3b 9 ай бұрын

5 or less.

@MarcelCox1 Жыл бұрын

Sorry, but I find your video disappointing for 3 reasons: 1. You did not specify in which domain you are solving, integers? real numbers? complex numbers?. You solved for complex, but you did not specify in advance. In math alympiads, problems are always specified more precisely. You omitted important information. This is an error often commited by a number of people doing this kind of videos, and I find it hugly annoying. 2. Your solution is correct but the problem can be solved more easily. In this kind of problems, there are often more efficient approaches than the brute force method. The colculations get easier when you realise that that x5+x5=(x+y)(x4-x3y+x2y2-xy3+y4) and continue from that. You then only need 4th and second powers of (x+y) which makes the fomulas a little bit simpler and thus less error prone. 3. This comment is mino: state your sources. Like xxx math olympiad problem from yyyy where xxx is the orangisation/country or whatever and dddd is the year. For poitn 3, I have the impression that many youtubers just copy the problems from other youtube videos as while there are so many interessing math olympiad problems, you oftend find the same problems over and over again.

@Костик-к2т Жыл бұрын

Х=1, У=2 ; У=2, Х=1.🙉🙉🙉🙉🙉

@calcnoon Жыл бұрын

・・これを日本の大学入試は 5分以内で全てやらにゃあかん。このスピードでは既に不合格。

@supermanusa4828 10 ай бұрын

X^5 + Y^5 # (X + Y)^5

@williammarshal4043 Жыл бұрын

Evaluate before doing all of your shit. If x+y = integers. and x⁵+y⁵= integer then x and y is integer then evaluate b²-4ac, if it's negative, stop. No point of continuing, i^odd integers. Will always has I as result.