Math Olympiad | A Nice Algebra Problem | How to solve for X and Y in this problem ?
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- Опубликовано: 2 окт 2024
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How to solve this math problem
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a very simple solution can be to use binomial therorem (x+y)^5 and then create the expansion in terms of x+y and xy, u will directly get the quadratic equation to calculate xy.
Agreed
I always feel like you complicate matters when it is actually unnecessary. Eg. X=4-Y ...(1) and XY=14...(2), just substitute (1) into (2) and solve the resultant quadratic equation. For the other case, XY=2...(3), So, substitute (1) into (3) and solve the quadratic equation. period.
Sure, i was expected the end of video at this point but suddenly he was in the middle of explanation.
I agree. 😅
Same thought! 😊
That what I thought as well.
Exactly 🙁
Another nice way to solve this equation, and I think a quicker one, is to substitute "x" with "2 + u" and "y" with "2 - u". This way, when we expand "(2 + u)^5 + (2 - u)^5 = 464" we get a biquadratic equation, which is easy to solve.
The equation is solved in 10 minutes not in 25 in the clip.
The elementary way to solve this is reducing the higher powers to lower powers by substitution.
Since x^5+y^5 = (x+y) (x^4-x^3y+x^2y^2-xy^3+y^4) => x^4-x^3y+x^2y^2-xy^3+y^4 = 116
At the same time (x+y)^4 = x^4+4x^3y+6x^2y^2+4xy^3+y^4 = 256
Subtracting gives 5x^3y+5x^2y^2+5xy^3 = 140 => xy (x²+xy+y²) = 28
We further reduce x²+xy+y² = (x+y)² - xy = 16-xy
This gives xy (16-xy) = 28 and this is a quadratic equation in xy with solutions 2 and 14
This gives 2 sets of equations (x+y=4, xy = 2) and (x+y=4, xy=14) which we can transform into quadratic equations by substitution
Those give solutions (2+v2, 2-v2) and (2+2i v10, 2-2i v10) where the latter exists in the complex field.
Even if you show this step by step, this can be done in 5 minutes. You are math literate but you take unnecessary long roads to the solution.
this is the real method that the question setter want the solvers to find out, not
the way in video
Thanks🔔
Correct
Why a degree 5 equation only have 4 sets of roots ? Is something missing ?
If you solve x^5+y^5 = 464 for one variable, you get five solutions because it's a fifth degree polynomial equation.
But this isn't a polynomial equation - it's a *system* of polynomial *equations*. Finding the number of solutions to a system of polynomial equations is much more difficult, because it's complicated by factors like the independence of equations and the possibility of infinitely many solutions (check out the wikipedia article for "system of polynomial equations").
In this particular case, we're lucky because the fifth degree polynomial x^5+y^5 is divisible by x+y so we just need to solve the fourth degree polynomial that results from the quotient, hence four solutions. (Also, normally we would need to prove the demoninator of our polynomial division is not zero, but here we are given x+y=4)
Once you have their sum and product you can write down the equation for _x_ and _y_ immediately, since
_t² - (x + y)t + xy = 0_
is satisfied when _t = x_ and when _t = y_
E.g. _x + y = 4, xy = 2_
_t² - 4t + 2 = 0_
with roots _2+√2, 2-√2_
which means that these are the respective values of _x_ and _y_ in either order.
This equation is aka Vieta's formula.
If f(n) = x^n + y^n then f(n+2) = f(1)*f(n+1) - xy*f(n) and f(0)=2
For this problem f(1) = 4, f(5) = 464, and we first need to solve for xy
Let xy = b then f(n+2) = 4*f(n+1) - b*f(n)
f(0) = 2
f(1) = 4
f(2) = 16 - 2b
f(3) = 64 - 12b
f(4) = 256 - 64b + 2b^2
f(5) = 1024 - 320b + 20b^2 = 464
20b^2 - 320b + 560 = 0
b^2 - 16b + 28 = 0
(b - 2)(b - 14) = 0
xy = 2 or xy = 14 each with 2 solutions
Как всё же хорошо знать формулы бинома Ньютона для (x + y)^3 и (x + y)^5, теорему Виета и метод подстановки! Сколько бумаги можно сэкономить!
Ага, видео - та ещё наркомания)
😊@@odinrossi3353
Да, точно!
А сколько ещё ошибок можно допустить во всей этой писанине!
Индус все усложнял
Если уравнение содержит х и у в пятой степени, то это уравнение пятой степени, и тогда всех возможных корней (включая комплексные) должно быть пять. А тут четыре
Esatto. Il ragazzo è bocciato
Hello . Please I need the solution to this equation. x,y are reals such that: x²-xy+y²=76 What is the value of x+y=?
I solved it I hoped nicely Allah Akbar 4 you ruclips.net/video/gQBFudhLdEs/видео.html
Why you are making it that much complicated
Good job! Well done!
Otherwise I understand that may be in different regions and countries it simply use another approach to Quadratic equations as: ax2 + bx + c = 0 to standard solution…., but was shown it is easy for simple and small numbers but otherwise it will be really difficult to determine the answers….
Thank you very much my dear friend
I have been away from math for 25 years. I was really shocked at how it came flooding back. Old brain but it is not dead yet!
Hay que practicar matemáticas todos los días. Tengo 72 añitos jaja y sigo practicando. Saludos desde Chiclayo Norte del Perú
Study this much simpler solution method. Substitute x=z+2 and y=-z+2 into the given equation and rearrange to (z+2)^5-(z-2)^5-464=0
Pascal's Triangle: 1 5 10 10 5 1; 2[5*2z^4+10*2^3z^2+2^5]-464=0; 20z^4+160z^2-400=0; z^4+8z^2-20=0
z^2=(-8±12)/2=2 or -10 and z=±√2 or ±i√10
x=z+2=2±√2 or 2±i√10 and y=4-x=2∓√2 or 2∓i√10
Thank you
this a long and delay process that take a lot of step, and a lot memorization require. This video show how to solve this equation with a garanteen answer.
Nice video! 👍 You should've written x1y1 and x2y2 at 11:51, because the way you wrote xy=14 and xy=2, it looks like you're saying that 14=2. 😁 Same thing for u. Should've been u1 and u2.
Наиболее краткое решение с верным ответом, при условии применения более простых методов - является верным!.... Здесь такого нет - решение излишне усложнено, из за этого более длинное - значит решение некорректно.... Можно было сразу использовать формулу а^n + b^n , при n = 5 .... Более того решив систему двух уравнений и в последствии квадратное уравнение получилось бы по человечески!
I've worked in mathematics for decades, I have no idea what cancel is as an operation. Everybody that uses it seems to confuse it with dividing out, or subtracting out. However: X/X=1, X-X=0. The reason most people have trouble with algebra is they're using the same word for two completely different operations.
There is no such operation as "cancel". Things either divide out, or they subtract out. You have to use the correct term if you're going to teach students what they're supposed to do.
Many thanks to Mr Burt Fetters, my college algebra teacher, who drilled this into the class, and finally made algebra an understandable system.
I agree with you that "cancel" can be confusing to beginner mathematicians who don't really know what's going on. So, when teaching elementary algebra, it should be avoided and instead the actual operations must be made explicit. As a shorthand for an elementary operation during a more elaborate exercise for algebra literates, I think it's fine.
On the mark!!When in doubt, cancel out is NOT MATHEMATICS but gibberish.
(x + y)² = x² + y² + 2xy → given: x + y = 4
16 = x² + y² + 2xy
(x + y)³ = (x + y)².(x + y)
(x + y)³ = (x² + 2xy + y²).(x + y)
(x + y)³ = x³ + x²y + 2x²y + 2xy² + xy² + y³
(x + y)³ = x³ + y³ + 3x²y + 3xy²
x³ + y³ = (x + y)³ - 3x²y - 3xy²
x³ + y³ = (x + y)³ - 3xy.(x + y) → given: x + y = 4
x³ + y³ = 64 - 12xy
(x + y)⁵ = (x + y)².(x + y)².(x + y)
(x + y)⁵ = (x² + 2xy + y²).(x² + 2xy + y²).(x + y)
(x + y)⁵ = (x⁴ + 2x³y + x²y² + 2x³y + 4x²y² + 2xy³ + x²y² + 2xy³ + y⁴).(x + y)
(x + y)⁵ = (x⁴ + 6x²y² + 4x³y + 4xy³ + y⁴).(x + y)
(x + y)⁵ = x⁵ + x⁴y + 6x³y² + 6x²y³ + 4x⁴y + 4x³y² + 4x²y³ + 4xy⁴ + xy⁴ + y⁵
(x + y)⁵ = x⁵ + y⁵ + 5x⁴y + 5xy⁴ + 10x³y² + 10x²y³
x⁵ + y⁵ = (x + y)⁵ - 5x⁴y - 5xy⁴ - 10x³y² - 10x²y³
x⁵ + y⁵ = (x + y)⁵ - 5xy.(x³ + y³) - 10x²y².(x + y) → given: x + y = 4
x⁵ + y⁵ = 1024 - 5xy.(x³ + y³) - 40x²y² → recall: x³ + y³ = 64 - 12xy
x⁵ + y⁵ = 1024 - 5xy.(64 - 12xy) - 40x²y²
x⁵ + y⁵ = 1024 - 320xy + 60x²y² - 40x²y²
x⁵ + y⁵ = 1024 - 320xy + 20x²y² → given: x⁵ + y⁵ = 464
464 = 1024 - 320xy + 20x²y²
20x²y² - 320xy + 560 = 0
x²y² - 16xy + 28 = 0 → let: z = xy
z² - 16z + 28 = 0
Δ = (- 16)² - (4 * 28) = 256 - 112 = 144 = 12²
z = (16 ± 12)/2
First case: z = 14 → recall: z = xy → xy = 14
Second case: z = 2 → recall: z = xy → xy = 2
Resume:
x + y = 4 ← this is the sum S
xy = 2 or xy = 14 ← this is the product P
So x & y are the solution of the following equation: a² - Sa + P = 0
First case: xy = 14 and (x + y) = 4
a² - 4a + 14 = 0
Δ = (- 4)² - (4 * 14) = 16 - 56 = - 40 = 40i²
a = (4 ± i√40)/2
a = (4 ± 2i√10)/2
a = 2 ± i√10
→ x = 2 + i√10
Recall: y = 4 - x
y = 4 - 2 - i√10
→ y = 2 - i√10
→ x = 2 - i√10
Recall: y = 4 - x
y = 4 - 2 + i√10
→ y = 2 + i√10
Second case: xy = 2 and (x + y) = 4
a² - 4a + 2 = 0
Δ = (- 4)² - (4 * 2) = 16 - 8 = 8
a = (4 ± √8)/2
a = (4 ± 2√2)/2
a = 2 ± √2
→ x = 2 + √2
Recall: y = 4 - x
y = 4 - 2 - √2
→ y = 2 - √2
→ x = 2 - √2
Recall: y = 4 - x
y = 4 - 2 + √2
→ y = 2 + √2
You could see that for x+y=4, xy=z, results will be symmetrical for each pair of rezults for both z.
Зачем же так сложно решать? Этот же ответ я получил через 5 минут.
Учителю надо бы поучиться умножать целые числа
Bài toán khá phức tạp phải áp dụng nhiều kiến thức toán học, cùng vài trang giấy mời giải được !
Так можно решать до бесконечности, а решить можно гораздо проще.
I more interested how many pages you used eventually
Những phép phân tích quá đơn giản mà diễn giải quá chi tiết làm mất thời gian và gây khó chịu quá
Total waste of paper
Despite this solution is very detailed, I think it is not entirely correct. Exponentiation of left and right equation parts, as well as multiplication of equation parts, is not equivalent transformation. x=-1 is not equivalent to x^2=(-1)^1, and even x=1 is not equivalent to x^3=1 in the field of complex numbers. If you added equations (3) and (4) to the initial system, it would be equivalent transformation of system as a whole. But in this case you should check every solution of your intermediate equation, whether it feats to both initial equations (1) and (2). And we all will see the practise of exponentiation to the power of 5 complex numbers and numbers with radicals. Or you should use fundamental theorem of algebra about polynomial complex root number, using transition to biquadratic equation, already mentioned in comments. But in this case may be it's worth just to use formula for biquadratic equation?
Take too long time, fail in an exam
It’s so long I think how to do in exam short period
Thanks
Это всё равно что лететь из Берлина в Париже через Пекин.
😂😂
You must use relation betwen th summ and product of the roots for equation of 2nd degrees.
Explained well sir
I have studied maths only upto 10th std. In +2 I've studied pure science
This solution which taught by you is very difficult to me. So, so hard sir.
How can I pick up it.🙄
😊😊
While it is good for brain 🧠 maintenance to think or practice logic, it is crucial to learn how to put problems in mathematical form. If your problem is not correctly described by a mathematical equation then result is wrong. Garbage in garage out is what they teach you in computer classes.
I literally think you complicated the procedure and got more video streaming budget. Sorry I shouldn’t think this way.
It's 46÷23 =2 then 4 divided by 2 thats 2 that's 2, 2s so it's 2to the power or 2 2square idk how to articulate how you say it properly in mathematics
nice solution i really love longer solutions it makes me feel great hahah
De vdd que fuma del weno, sinceramente veo muchos errores matematicos
I think you'd get more positive feedback if you offered this as an alternative solution and do quadratic in method 1.
Very bad approach to solve this problem
IS a big fail cancel (x+y)² with the squere root , because this is the a absolute value definition. So is wrong !!!
Siempre te complicás Ecotú querido. Sustitución x=4-y, lo sustituyes en la 2a ecuación. Mirá que fácil. No te compliques tanto Ecotú.
too complicated .. xy=P and x+y=S you can solve by x^2 -Sx +P = 0 ...
When we meet x+y=a and xy=b, we better say x,y are the solutions of t^2-at-b=0. Though in the video (x-y)^2=8 is changed into x-y=±2√2, it should be |x-y|=2√2 and will need case treatment. We may omit them with proposed explanation.
t^2-at+b
Should be:
t^2 - at + b = 0
why not try brute force method : X = 3.4142, : y = 0.585799. Thanks
like your pen, whts the type name?
El ejercicio mas largo de la historia, innecesario
When I asked Chatgpt
It is said that the x ≈ is 2.532 and the y ≈ is about 1.468.
Of the solutions two were imaginary and two were trivial, but since one of the imaginary solutions was also trivial the is in only one non-imaginary non-trivial solution.
X+Y ruclips.net/video/Icht6ynLO7I/видео.html
If I was reviewing your work, I would give you a low grade for not using a simpler solution.
x5,y5=?
1. \(x = 2 - \sqrt{2}, y = \sqrt{2} + 2\)
2. \(x = 2 + \sqrt{2}, y = 2 - \sqrt{2}\)
还有两个复数解，通常情况下我们只考虑实数解。
A symple problem made complicated and much more the time 25 mins !
A good mathematician does not make a good teacher. I suggest using a logic line on the right of your equations so that you can explain their thought processes.
Лучше бы на своем родном языке говорил. На арабском лучше всего.
Çok önemliydi sanki
Trực tiếp mũ 5 phương trình x+y=4 cho ra cái x^5 và y^5 luôn
Good job mate!
Thank you very much!!!
対称式なので、ｘｙを求めることが鍵だとは分かったが、その求め方が自分のやり方と違ってスマートなのに感心しました。あとｘｙを導き出した後の最後の解の求め方は２次方程式の解と係数の関係を自分は使った。
Interesting problem, interesting solution technique. BUT you belabor and agonize over basic arithmetic, but leap over the error-prone operations such as collection of terms. Your presentation could be so much more helpful if you show the error-prone steps, but move quickly through the truly elementary stuff.
Followup: Each step you show should be as much as can be verified by quick inspection, neither more nor less. Anyone taking this problem on already has elementary algebra. The magic of this solution lies in generating formulas that allow the repeated substitution of known values. That's the magic, and the structure of that magic should be foremost in the presentation.
Use vietta types
This is a joke!
Something bothers me. Intuitively, I would have thought that a 5 degrees équation should have 5 couples of solutions in C, no? Just like a 5 degrees equation of x has 5 solutions in C. I would guess that I'm wrong somewhere, but i dont know where, and if i'm not where ois the fifth solution ?
There's an extra constraint x+y=4. This reduces the solution space with (at least) one degree.
Nice pen
Using Powers And roots to equation have some rules And leads to reducing interval of usable numbers.
You have unusual english pronounce)
What a long ending !!!
At the end, once you have xy=14 and x+4 = 4
x(4-x) = 14
x^2 - 4x + 14 = 0
x = (4 ± √-40)/2 , or x = 2 ± √10 i , same thing for the other case , so you your 2 x's , easy to get the y
Beautiful work.
Thank you very much!!!
How many pages you used 😊eventually
Can you help me compute the flux integral of the surface S using the divergence theorem if the vector field F = yj and S is a closed vertical cylinder of height 2, with its base a circle of radius 1, on the xy-plane, centered at the origin ?
А зачем так сложно то? ))) Пришла пора попробовать себя в физике или химии
x^5+y^5＝464なので
5乗の場合 解が5個存在しないとおかしいので
解が4個は不正解です
ヒントだけ書いときますね
もう一個の解は2を使いません ｘかｙにマイナスがあるため 結構大きな数字が入ります
I wrote this in another comment, but the reason we have 4 solutions is because this is a "system of polynomial equations" instead of a polynomial equation. In this particular system, we can divide x^5+y^5 by x+y to obtain a 4th degree equation, which is what is being solved. This division is valid because x+y=4≠0
Has lost his way.
完全看蒙了，如果是考试，估计半个小时做不完一道题呀😂
Haven't you studied about a discriminant of a quadratic equation? 8th or 9th grade in the school. The problem is much easier to solve, not in 25 minutes.
Dude hates paper.
Too complicated.
Don't solve more Mathe questions
Why ?
12:45 es obvio pero como se te ocurrió
Жесть. Это где ж так учат?
Bad quiestion
x = ~.58
y = ~3.42
Quá phức tạp
Sastisfhu
Thank you 👍👍
I find it really poor practice to use "x" for both a variable and a multiplication symbol in the same equations. So much opportunity for confusion, and it really slows down the scanning of a line.
Terrible headache
Μπράβο!!!
Я тебя 20минут смотрела, чтоб узнать. Ты конкретно определишь чему равны x и y? Просто там а первом уравнении видно, что один равен 1, а второй 3. Просто можно числа поменять. От этого не изменится решение ни одного ни второго уравнения
Но только 1^5 + 3^5 не равно 464
Thank you❤❤
Bruh . X=4-y …
X^5 + Y^5 = 464
:)
Х=1, у=3(х=3 у=1)
Miért nem használod a másodfokú egyenlet képletét?
❤❤❤❤
算完天都亮了，下一位
Can someone explain to me at 13:00 when he just puts in (x+y)^2 + (x-y)^2= 4xy, where do the (x-y)^2 and 4xy come from? It feels like he just adds them in out of nowhere, though he says it’s an algebraic rule.
it is （x+y)^2 - (x-y)^2=4xy