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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.
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.
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.
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!
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
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 ?
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.
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)
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….
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
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.
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.
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.
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.
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.
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
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.
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
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 ?
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 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.
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:-)
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.
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.
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?
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.
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
