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.
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.
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.
Yeah, I hate how these equations are made way more complicated than they need to be. Some I can solve in my head in 10 seconds. This one, needs paper@@IvanPetrov-rs1kq
I did it this way, before actually watching the lengthy video. After binomial expansion and cancelling, it reduces to u^4+8u-20=0. then either u^2=2 or u^2=-10. For real solutions, t^2=2, thus t=+-sqrt(2). The real value for the (x,y) pair is (2+sqrt(2), 2-sqrt(2)). The complex pair would be (2+i*sqrt(10),2-i*sqrt(10)). Much quicker.
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.
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.
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 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.
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
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.
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….
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.🙄
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.
Гораздо проще решать методом итераций: подставляем 2, первое выражение попадаем в точку а во втором при прямой подстановке получаем значение равное 64. Артнллеристы скажут, что до цели большой недолет. Подставляем в оба выражения Х=3, в обоих выражениях получаем большой перелет. Во втором выражении Х=648, поэтому взяв эти выражения в вилку(как говорят артеллериств) будем подбирать промедуточное число на отрезке 2 и 3.
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)
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.
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.
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.
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
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.
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 ?
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
The solution is straight forward for school students: y=4-x => x^5+(4-x)^5=464 (x^2-4x+2)(x^2-4x+14)=0. For real solutions , (x^2-4x+2)=0 (x-2)^2-2=0,. Alternatively, Vieta's formula and symmetry.
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 ?
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?
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.
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.
Buenas tardes, hermosa explicación. Y, como sostengo, las matemáticas, sus reglas y sus números SON EL IDIOMA...no importa la lengua en que se explique. Gracias
Кстати, у Вас ценное замечание на счёт проверки корней уравнений. 👍 При возведении частей уравнения в степень, при замене переменной (что в данном решении встречалось) могут появиться "посторонние корни", которые не являются корнями исходного уравнения. Так что проверка обязательна и без неё решение не является полным:-)
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.
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.
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
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
This is no different from what the poster did. Same process of reducing powers.
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
binomial theorem is the fastest way to solve this thing.
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
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.
Yeah, I hate how these equations are made way more complicated than they need to be. Some I can solve in my head in 10 seconds. This one, needs paper@@IvanPetrov-rs1kq
I did it this way, before actually watching the lengthy video. After binomial expansion and cancelling, it reduces to u^4+8u-20=0. then either u^2=2 or u^2=-10. For real solutions, t^2=2, thus t=+-sqrt(2). The real value for the (x,y) pair is (2+sqrt(2), 2-sqrt(2)). The complex pair would be (2+i*sqrt(10),2-i*sqrt(10)). Much quicker.
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.
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.
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 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 🙁
Beautiful work.
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。
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
You could see that for x+y=4, xy=z, results will be symmetrical for each pair of rezults for both z.
Thank you very much my dear friend
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
Using Powers And roots to equation have some rules And leads to reducing interval of usable numbers.
Μπράβο!!!
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.
Так можно решать до бесконечности, а решить можно гораздо проще.
nice solution i really love longer solutions it makes me feel great hahah
Nice problem. Nice job in solving it.
Thank you very much!!!
Nice!
Thank you! Cheers!
x + y = 4
x⁵ + y⁵ = 464
Let x = 2 + n. So
y = 4 - x
y = 4 - (2 + n)
y = 2 - n
So the second equation becomes
(2 + n)⁵ + (2 - n)⁵ = 464
Expand and simplify the powers:
(2 + n)⁵ = 2⁵ + 5*2⁴n + 10*2³n² + 10*2²n³ + 5*2n⁴ + n⁵
(2 - n)⁵ = 2⁵ - 5*2⁴n + 10*2³n² - 10*2²n³ + 5*2n⁴ - n⁵
(2 + n)⁵ + (2 - n)⁵ = 2⁵ + 2⁵ + 10*2³n² + 10*2³n² + 5*2n⁴ + 5*2n⁴
(2 + n)⁵ + (2 - n)⁵ = 32 + 32 + 80n² + 80n² + 10n⁴ + 10n⁴
(2 + n)⁵ + (2 - n)⁵ = 64 + 160n² + 20n⁴
So you get to solve
64 + 160n² + 20n⁴ = 464
20n⁴ + 160n² + 64 - 464 = 0
20n⁴ + 160n² + 400 = 0
20(n⁴ + 80n² - 20) = 0
n⁴ + 80n² - 20 = 0
(n²)² + 80n² - 20 = 0
s² + 80s - 20 = 0
Solve the quadratic equation for s = n² ≥ 0:
s = (- 8 ± √(8² - 4*1*(-20)))/(2*1)
s = (-8 ± √(64 + 80))/2
s = (-8 ± √144)/2
s = (-8 ± 12/2)
Since s = n² > 0, reject the negative value for s:
s = (-8 + 12)/2
s = 4/2
s = 2
Substitute back s = n²:
n² = 2
n = ±√2
Therefore,
x = 2 + n
x = 2 ± √2
y = 2 - n
y = 2 ∓ √2
Solution:
(x, y) ∈ {(2 + √2, 2 - √2), (2 - √2, 2 + √2)}
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❤❤
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。
You must use relation betwen th summ and product of the roots for equation of 2nd degrees.
It’s so long I think how to do in exam short period
Thanks
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.🙄
Good job mate!
Thank you very much!!!
❤❤❤❤
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
Гораздо проще решать методом итераций: подставляем 2, первое выражение попадаем в точку а во втором при прямой подстановке получаем значение равное 64. Артнллеристы скажут, что до цели большой недолет. Подставляем в оба выражения Х=3, в обоих выражениях получаем большой перелет. Во втором выражении Х=648, поэтому взяв эти выражения в вилку(как говорят артеллериств) будем подбирать промедуточное число на отрезке 2 и 3.
So you will get only the real answers!
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)
I think you'd get more positive feedback if you offered this as an alternative solution and do quadratic in method 1.
Bài toán khá phức tạp phải áp dụng nhiều kiến thức toán học, cùng vài trang giấy mời giải được !
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.
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.
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
like your pen, whts the type name?
Nice
Thank you 👍👍👍
Nice pen
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
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 ua-cam.com/video/gQBFudhLdEs/v-deo.html
I like how you always say "bracket" but always put a "parenthesis"❤
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.
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
Зачем же так сложно решать? Этот же ответ я получил через 5 минут.
Quá phức tạp
😊😊
How many pages you used 😊eventually
4 da matina, parei de assistir na hora que ela mudou de folha, poucas ideia pra Y menos ainda pra X
I more interested how many pages you used eventually
Miért nem használod a másodfokú egyenlet képletét?
I am as dumb as I started... If anything I am dumber...
坐標內有i，是什麼意思？
是什麼意思？
將-1開根號之值，是個虛數，用i代表。
root -1 = i
完全看蒙了，如果是考试，估计半个小时做不完一道题呀😂
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
A symple problem made complicated and much more the time 25 mins !
一般聯立方程要的是實數解！
x = ~.58
y = ~3.42
Trực tiếp mũ 5 phương trình x+y=4 cho ra cái x^5 và y^5 luôn
too complicated .. xy=P and x+y=S you can solve by x^2 -Sx +P = 0 ...
The solution is straight forward for school students: y=4-x => x^5+(4-x)^5=464 (x^2-4x+2)(x^2-4x+14)=0. For real solutions , (x^2-4x+2)=0 (x-2)^2-2=0,. Alternatively, Vieta's formula and symmetry.
Учителю надо бы поучиться умножать целые числа
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.
Just divide by five and split it in half
Х=1, у=3(х=3 у=1)
Siempre te complicás Ecotú querido. Sustitución x=4-y, lo sustituyes en la 2a ecuación. Mirá que fácil. No te compliques tanto Ecotú.
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?
When I asked Chatgpt
It is said that the x ≈ is 2.532 and the y ≈ is about 1.468.
Why only 4 solutions for quintic equation?
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.
X^5 + Y^5 = 464
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.
why not try brute force method : X = 3.4142, : y = 0.585799. Thanks
Buenas tardes, hermosa explicación. Y, como sostengo, las matemáticas, sus reglas y sus números SON EL IDIOMA...no importa la lengua en que se explique. Gracias
...io vado al cinema..😢
Forse questi video sono rlivolti a chi non ha dimestichezza con la matematica e si sceglie un metodo più lungo,ma più semplice da comprendere...
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
12:45 es obvio pero como se te ocurrió
Это всё равно что лететь из Берлина в Париже через Пекин.
😂😂
x5,y5=?
Подставлял кто-то эти решения в уравнение?У меня не выходит(выходит на первое).
Да, подставлял.
Обращаются в верные равенства.
Все четыре пары являются корнями системы
Кстати, у Вас ценное замечание на счёт проверки корней уравнений. 👍
При возведении частей уравнения в степень, при замене переменной (что в данном решении встречалось) могут появиться "посторонние корни", которые не являются корнями исходного уравнения.
Так что проверка обязательна и без неё решение не является полным:-)
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.
Why you are making it that much complicated
X .y tinh bình phương lập phương trình vv. Quy đồng mẫu số và rút gọn phân số vv. OK bạn vvv.
IS a big fail cancel (x+y)² with the squere root , because this is the a absolute value definition. So is wrong !!!
Ať time 8.36 u²-6u²!=5u² you
forget minus
He's moving everything to the right side of the equation, so 5u^2 is correct.
:)
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.
Những phép phân tích quá đơn giản mà diễn giải quá chi tiết làm mất thời gian và gây khó chịu quá
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
Use vietta types
算完天都亮了，下一位
If I was reviewing your work, I would give you a low grade for not using a simpler solution.
Congratulations. You found the way to hate maths.
Fo u know math or u are writing mohabharat