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).
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.
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.
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.
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.
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.
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.
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.
With integers as condition, it’s a simpler exercise. Otherwise there should be 3 sets of solutions. Can do another round of substitutions u=a+b and v = ab
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.
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
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.
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.
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)
@@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 !
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.
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).
At time of 6:20, I think that there is a mistake. You wrote that "a" should be +/- 4. Did you mean+/- 2, instead?
he changed it at 9:18
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.
One of the most entertaining people on the internet. Thank you so much for providing us with these maths problems and worked solutions :) ♥
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.
Two in one: Math lesson and English one as well:)) Thank you, sir!
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.
I was thinking exactly the same idea with you. We cannot assume the 6th roots of integer must be an integer.
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.
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.
I have never thought that ı will be having fun while watching math videos. You are amazing keep going man!
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.
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.
Dude 😂
Ypur videos bring me so much calm and peace. Thank you so much for everything you do
Surd[x,3]+Surd[y,3]=13 Sqrt[x]+Sqrt[y]=35 (x,y)=(64,729),(729,64) It’s in my head.
With integers as condition, it’s a simpler exercise. Otherwise there should be 3 sets of solutions. Can do another round of substitutions u=a+b and v = ab
Beautiful teaching
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.
You could we were given a (h)int. (:
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
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.
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
(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
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.
Awesome substitution ❤❤😊
Put x=a^6,y=b^6 and solve.
(a,b)=(3,2) or (2,3)
(x,y)=*729,64) or (64,729).
you didn't need the ± because from the square roots, x and y should be ≥ 0
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)
@@arthurmorgan3970 mate , if u plug in negative values in the equation it wont work from what i think . pls correct me if im wrong
@@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 !
@@arthurmorgan3970 yes but why and how do u know sir
@@monkeblazer3154cause most of the brilliant mind people on American videos are indians.
@@simonghostriley9657 oh ic
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.
In which book you solve this question please tell the book name
Rs aggarwal
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.
Can you do a video on functional differential equations ?
Fantastic!!
square root of a positive number is positive. You don't need to find a or b , but X
Proper work
Got the answer but the difficulty is in bringing it to the simplest form.
Can you please make a video on log and natural log
👍🧠
At the end is a *typo, it is not b but y. 😊😊
A sqr of a positive real number ist defined as positive.
10:17 i dont think you meant to write b there 😂
Are you related to Omar Epps you could be brothers lol.
Hello