System of Radical Equations

[@florianbasier]

5:47 but you don't know that a and b are integers. x and y are, but a and b are their 6-th roots so don't have to be. You proved that your solution worked but not that it is the only one.

[@Christian_Martel]

Exactly.

[@markos635567]

Immediate thought: Since both results are integers, all roots should evaluate to integers (not trivially true but a good first guess), thus both answers must be raised to the sixth power (the least common multiple). Which gives us the obvious 4+9 on the first, in other words x=2^6 and y=3^6. Which then checks out on the bottom (8+27).

[@domanicmarcus2176]

At time of 6:20, I think that there is a mistake. You wrote that "a" should be +/- 4. Did you mean+/- 2, instead?

[@clemberube6681]

he changed it at 9:18

[@anatoliy3323]

Two in one: Math lesson and English one as well:)) Thank you, sir!

[@pietergeerkens6324]

Accepting unrestricted answers: from 2:24 substitute again with s = u + v and p = u * v. Then the expansions of the s cubed and s squared give s^2 = 13 + 2p s^3 = 35 + 3sp. From solutions for s and p, use Viete's Rules to obtain u,v and then x,y in any domain, whether N, Z, R, or C.

[@ronaldlincon6679]

One of the most entertaining people on the internet. Thank you so much for providing us with these maths problems and worked solutions :) ♥

[@darklightmotion5534]

Ypur videos bring me so much calm and peace. Thank you so much for everything you do

[@elidepp3553]

Талантливый. Способ написания, жесты, мимика, работа голосом - всё великолепно проработано. Отличная работа! И кепка шикарная. Но у меня лучше)

[@gokalpgorduk7685]

I have never thought that ı will be having fun while watching math videos. You are amazing keep going man!

[@andrewjames6676]

Beautiful teaching

[@Etothe2iPi]

The equations are symmetrical in x and y. So you just have to calculate one solution and say in the end that you can swap the two values, because of symmetry.

[@duckyoutube6318]

Brave man wearing that kind of shirt while working with a chalkboard. All it takes is one small streak, one tiny bump, and boom! You're shirt is covered in chalk. Then you gotta explain to everyone that tells you about the smudge on your shirt how you got it. And not only that but now the cute girl serving coffee down the street thinks you hang drywall for a living and now expects you to be handy around the house. Its a lose/lose situation.

[@JatPhenshllem]

Dude 😂

[@samueltso1291]

I have a quicker way. 35 = 27 + 8 = 3^3 + 2^3 13 = 9 + 4 = 3^2 + 2^2 Put m = x^(1/6) and n = y^(1/6) The equations become x^(1/3) + y^(1/3) = 13 ==> m^2 + n^2 = 3^2 + 2^2 --------- (1) x^(1/2) + y^(1/2) = 35 ==> m^3 + n^3 = 3^3 + 2^3 ----------(2) From (1) and (2), we can conclude that m = 3 and n = 2 or m = 2 and n = 3 m = 3 ==> x^(1/6) = 3 ==> x = 729 n = 2 ==> y^(1/6) = 2 ==> y = 64 Or x = 64 and y = 729.

[@alibhukoo5400]

Awesome substitution ❤❤😊

[@cicik57]

it can be solved in mind really fast. in integers it is the same as x² + y² = 13, x³ + y³ = 35. The first equasion gives obvious solutions x = 2, y = 3. Check: 8 + 27 = 35. So the original x = 2^6 and y = 3^6 or swap values.

[@xyz9250]

With integers as condition, it’s a simpler exercise. I tried without that assumption, did another round of substitutions u=a+b and v = ab, and solved it by factoring. it turns out (2,3) are still the only valid solution for a and b.

[@aashsyed1277]

5:01 the square root is always positive so we don't need the plus or minus

[@amtep]

You made the assumption that a and b are integers when you restricted them to perfect squares. But I don’t think that follows. x and y have to be integers but their 6th roots aren't necessarily integers.

[@하태욱-b2x]

I was thinking exactly the same idea with you. We cannot assume the 6th roots of integer must be an integer.

[@benkahtan6802]

Yeah. With some manipulation, you can find from the first equation that xy must be a cube, and you can find from the second equation that x must be a square and y must be a square. Since x and y must be squares, all of their prime factors must have even powers. For xy to be a cube, all of its prime factors must have powers divisible by 3. You can't have a factor p have power 1 in x and 2 in y, since then x won't be square. So all prime factors of x must have powers divisible by 2 and by 3, that is, divisible by 6, and the same goes for y.

[@RyanLewis-Johnson-wq6xs]

Surd[x,3]+Surd[y,3]=13 Sqrt[x]+Sqrt[y]=35 (x,y)=(64,729),(729,64) It’s in my head.

[@michaelkiniklis9150]

This is a system of two equations with two unknown entities, which means that there is a solution that can always be found without requiring any additional restriction. The requirement that the solutions be integers is an additional restriction that can only be true by chance, which is the case in this system.

[@johanmedioni7856]

You could we were given a (h)int. (:

[@pietergeerkens6324]

There are three (pairs) of solutions: - A symmetrical pair in N; - A symmetrical pair in R; and - A complex conjugate pair in C. See my comment above. To find all of them, substitute s = u + v, p = uv, and from the resulting solution set apply Viete's Rules

[@skwbusaidi]

We can assume that a &b are non nagative integers from the beginning because we interested in x and y which they are non nagative integers

[@abdulhusseinalsultani9222]

Since the right side is an integer so the cubic roots in the left side must be integer numbers and also the nearest number to 13 is 9 so 13_9=4 and this means that 64^1/2+729^1/2=8+27=35 answer

[@kimutaiboit8516]

I made u⁶=x and t⁶=y From there I made these u²+t²=13 u³ + t³ = 35 From there it was easy peazy. Great minds think alike.

[@reyadhaloraibi3387]

Fantastic!!

[@monkeblazer3154]

I took cube root of x and cube root of y to be a and b respectively . i took root a = c and root b = d , from their using the equations i got cd as 6 or 8 where only 6 is possible due to the constraint of the equation , so c + d will be 5 , but c^3 + d^3 must give 35 so only possible when they are 2 and 3 respectively . so a = 9 or 4 , b = 9 or 4 , therefore , x ,y (64 , 729 or otherwise)

[@monkeblazer3154]

@@arthurmorgan3970 mate , if u plug in negative values in the equation it wont work from what i think . pls correct me if im wrong

[@monkeblazer3154]

@@arthurmorgan3970 im basically eliminating some possibilities to pull out the correct answer due to the constraints the equations offer . Is there an alternative to solve it ? Unfortunately i cant send a photo but maybe the photo would make it clear !

[@monkeblazer3154]

@@arthurmorgan3970 yes but why and how do u know sir

[@simonghostriley9657]

​@@monkeblazer3154cause most of the brilliant mind people on American videos are indians.

[@monkeblazer3154]

@@simonghostriley9657 oh ic

[@AbouTaim-Lille]

U put x= y⁶. To simplify ig a little but and turn it into a polynomial. With a condition x>0. You should get an equation of 6th defree which since it is above 4th degree in general has no explicit formula for the solution.

[@adammohamed5256]

Proper work

[@MARTINWERDER]

Alternative solution method: to make a list with the pairs, whose sum of the cubic roots of n^1/3 + u^1/3 = 13 x x^1/2 x^1/3 y^1/3 y^1/2 y 216 not integer 6 7 not integer 343 125 not integer 5 8 not integer 512 64 8 4 9 27 729 square root 27 + 8 = 35 27 not integer 3 10 not integer 1000 8 not integer 2 11 not integer 1331 By chance, I started in the middle of the list with the pair 4 / 9 (x, y ) = [ (729, 64), (64, 729) ] and immediately had the result. But if you start the list at the beginning, you can see that all the other pairs don't have an integer square root of x or/and y.

[@chengkaigoh5101]

Can you do a video on functional differential equations ?

[@anestismoutafidis4575]

(x^1/3+y^1/3)+(x^1/2+y^1/2)=13+35 73^1/2=8,54; 73^1/3=4,17 [(8,54+4,17=12,7); -0,3 of 13] 700^1/2=26,45; 700^1/3=8,87; [(26,45+8,87=35,32) + 0,3 of 35] We can try ro put in: (700)^1/3 + (73)^1/3=13,0; (700)^1/2 + (73)^1/2=35,0; x=700; y=73

[@AshrafAli-e1l]

I am going to join the Pakistan Mathematics Olympics which is doing at first time in Pakistan

[@georgesbv1]

square root of a positive number is positive. You don't need to find a or b , but X

[@anoopyadav5617]

In which book you solve this question please tell the book name

[@learnwithdk5533]

Rs aggarwal

[@gnanadesikansenthilnathan6750]

Got the answer but the difficulty is in bringing it to the simplest form.

[@Thampuran-o9o]

👍👍👍

[@SrisailamNavuluri]

Put x=a^6,y=b^6 and solve. (a,b)=(3,2) or (2,3) (x,y)=*729,64) or (64,729).

[@Jianlong-xp5li]

Can you please make a video on log and natural log

[@Christian_Martel]

At the step a^2 +b^2 = 13, the only fact we can say is that a^2 and b^2 are positive and smaller than the square root of 13. We didn't prove that a^2 and b^2 are integers. You can't say that a^2 and b^2 are perfect squares.

[@uwelinzbauer3973]

👍🧠

[@kurtecaranum3023]

you didn't need the ± because from the square roots, x and y should be ≥ 0

[@thomaslangbein297]

A sqr of a positive real number ist defined as positive.

[@aravindhvijayanandan3010]

At the end is a *typo, it is not b but y. 😊😊

[@nasrullahhusnan2289]

Let u=x^⅙ and v=y^⅙ u²=x^⅓ and v²=y^⅓ u³=x^½ and v³=y^½ u²+v²=13 --> 2²+3²=13 u³+b³=35 --> 2³+3³=35 u=2 and v=3 --> x=u⁶ --> x=2⁶=64 y=v⁶ --> y=3⁶=729 (x,y)={64 729),(729,64)} as the equations are symmetrical.

[@donwald3436]

Are you related to Omar Epps you could be brothers lol.

[@creativename.]

10:17 i dont think you meant to write b there 😂

[@paulortega5317]

aw, let's get all the solutions let x = u⁶ and y = v⁶ let f(n) = uⁿ + vⁿ let f(1) = u + v = a let uv = b f(n+2) = f(1)•f(n+1) - (uv)•f(n) = a•f(n+1) - b•f(n) f(0) = 2 f(1) = a f(2) = a² - 2b = 13 f(3) = a³ - 3ab = 35 Reduce to just "a" a³ - 39a - 70 = 0 (a - 5)(a - 2)(a + 7) = 0 a = 5, 2, -7 b = (a² - 13)/2 (a,b) = (5,6), (2,-9/2),(-7,18) u + v = a and uv = b u + b/u = a u² - au + b = 0 u =[a ± √(a² - 4b)]/2 ❶ (a,b) = (5,6) u = [5 ± √(25 - 24)]/2 = (5 ± 1)/2 (u,v) = (3,2),(2,3) (x,y) = (729,64),(64,729) ❷ (a,b) = (2,-9/2) u = [2 ± √(4 + 18)]/2 = 1 ± √(22)/2 (u,v) = ((1 + √(22)/2),(1 - √(22)/2)),((1 - √(22)/2),(1 + √(22)/2)) (x,y) = ((1 + √(22)/2)⁶,(1 - √(22)/2)⁶),((1 - √(22)/2)⁶,(1 + √(22)/2)⁶) ❸ (a,b) = (-7,18) u = [-7 ± √(49 - 72)]/2 = (-7 ± √(23)i)/2 (u,v) = ((-7 + √(23)i)/2,(-7 - √(23)i)/2),((-7 - √(23)i)/2,(-7 +√(23)i)/2) (x,y) = (((-7 + √(23)i)/2)⁶,((-7 - √(23)i)/2)⁶),((-7 - √(23)i)/2)⁶,(-7 +√(23)i)/2)⁶)

[@Jianlong-xp5li]

Hello