In this video, I solved a system of radical equations by applying a change of variables using appropriate power choice
Пікірлер: 46
@markos6355677 ай бұрын
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).
@anatoliy33237 ай бұрын
Two in one: Math lesson and English one as well:)) Thank you, sir!
@ronaldlincon66797 ай бұрын
One of the most entertaining people on the internet. Thank you so much for providing us with these maths problems and worked solutions :) ♥
@darklightmotion55346 ай бұрын
Ypur videos bring me so much calm and peace. Thank you so much for everything you do
@andrewjames66767 ай бұрын
Beautiful teaching
@domanicmarcus21767 ай бұрын
At time of 6:20, I think that there is a mistake. You wrote that "a" should be +/- 4. Did you mean+/- 2, instead?
@clemberube66817 ай бұрын
he changed it at 9:18
@alibhukoo54007 ай бұрын
Awesome substitution ❤❤😊
@pietergeerkens63247 ай бұрын
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.
@reyadhaloraibi33877 ай бұрын
Fantastic!!
@duckyoutube63187 ай бұрын
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.
@JatPhenshllem7 ай бұрын
Dude 😂
@Etothe2iPi7 ай бұрын
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.
@gokalpgorduk76857 ай бұрын
I have never thought that ı will be having fun while watching math videos. You are amazing keep going man!
@adammohamed52567 ай бұрын
Proper work
@xyz92503 ай бұрын
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
@michaelkiniklis91507 ай бұрын
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.
@johanmedioni78567 ай бұрын
You could we were given a (h)int. (:
@pietergeerkens63247 ай бұрын
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
@uwelinzbauer39737 ай бұрын
👍🧠
@chengkaigoh51017 ай бұрын
Can you do a video on functional differential equations ?
@monkeblazer31547 ай бұрын
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)
@monkeblazer31547 ай бұрын
@@arthurmorgan3970 mate , if u plug in negative values in the equation it wont work from what i think . pls correct me if im wrong
@monkeblazer31547 ай бұрын
@@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 !
@monkeblazer31547 ай бұрын
@@arthurmorgan3970 yes but why and how do u know sir
@simonghostriley96577 ай бұрын
@@monkeblazer3154cause most of the brilliant mind people on American videos are indians.
@monkeblazer31547 ай бұрын
@@simonghostriley9657 oh ic
@kimutaiboit85167 ай бұрын
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.
@skwbusaidi2 ай бұрын
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
@AbouTaim-Lille7 ай бұрын
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.
@georgesbv17 ай бұрын
square root of a positive number is positive. You don't need to find a or b , but X
@Jianlong-xp5li7 ай бұрын
Can you please make a video on log and natural log
@anestismoutafidis4575Ай бұрын
(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
@gnanadesikansenthilnathan675021 күн бұрын
Got the answer but the difficulty is in bringing it to the simplest form.
@richardbraakman74697 ай бұрын
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.
@user-xw4ul9xr9uАй бұрын
I was thinking exactly the same idea with you. We cannot assume the 6th roots of integer must be an integer.
@thomaslangbein2977 ай бұрын
A sqr of a positive real number ist defined as positive.
@florianbasierАй бұрын
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.
@kurtecaranum30477 ай бұрын
you didn't need the ± because from the square roots, x and y should be ≥ 0
@anoopyadav56177 ай бұрын
In which book you solve this question please tell the book name
@learnwithdk55337 ай бұрын
Rs aggarwal
@nasrullahhusnan2289Ай бұрын
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.
@donwald34367 ай бұрын
Are you related to Omar Epps you could be brothers lol.