An Interesting Radical Equation | Math Olympiad Training

An Interesting Radical Equation | Math Olympiad Training
Welcome to infyGyan! In this video, we present an incredible radical equation that will enhance our algebraic skills and challenge our mathematical thinking. This problem is perfect for those preparing for Math Olympiad or anyone who loves tackling tough math problems.
Join us as we break down the steps to solve this intriguing equation, and see if you can find the solution on your own. Don't forget to leave your answers and approach in the comments below!
If you enjoy solving complex math problems, make sure to subscribe to our channel and hit the notification bell for more exciting math challenges.
Additional Resources:
• A Tricky Radical Equat...
• A Nice Radical Equatio...
• Can You Solve This Rad...
• A Nice Factorial Equat...
#algebra #math #problemsolving #radicalequation #matholympiad #education #tutorial #matholympiadpreparation #matholympiadtraining
🔔 Don't forget to subscribe and hit the bell icon to stay updated with our latest math challenges and tutorials. If you enjoy the video, please give it a thumbs up and share it with your friends!
💬 Comment below if you have any questions or if you were able to solve the problem. We love hearing from our viewers!
Thank you for watching and happy problem-solving!

[@johnlv12]

excellent solution

[@sarantis40kalaitzis48]

I prefer your's elegent solution by using t=a*b because is easier than mine with Binomial theorem, which can't be applyied when we have even exrponents like a^4+b^4 or a^6+b^6. Only odd a^n - +b^n.

[@mathpro926]

nice solution

[@paulortega5317]

At 5:16, a different way to determine the value for "t"
let f(n) = aⁿ + bⁿ
let a + b = s
let ab = t
then f(n+2) = s*f(n+1) - t*f(n)
f(0) = 2
f(1) = s
f(2) = s² - 2t
f(3) = s³ - 3st
f(4) = s⁴ - 4s²t + 2t²
f(5) = s⁵ - 5s³t + 5st² = 32
we had already determined s = 2 and f(5) = 32
32 = 32 - 40t + 10t²
t² - 4t = 0
t = 0, t = 4
etc.

[@jaggisaram4914]

8

[@tejpalsingh366]

x#

[@user-kp2rd5qv8g]

Let 8+√x=a^5 and 8-√x=b^5. Let a+b =u = 16^1/5 and ab=v. Using (a+b)^5 = a^5+b^5+5ab(a^3+b^3) + 10a^2b^2(a+b), we have uv(u^2-v)=0. Iv uv=0 and since u is not zero, v=o > 64-x=0 > x=64. If u^2-v=0 > 16^2/5 = (64-x)^1/5 > x = -192. But x must be positive. So, x=64.

[@ΜαργαριταΚαναρη]

Ειναι χ>0.μετα απο πραξεις χ=64

[@user-kt1dm9jz5t]

X=64, 64-2(8)^1/5

[@kelkarbharat9161]

The way youu write 5 in fifth root is incorrect.

[@dorkmania]

Dividing equation by (8 - √x)^⅕
((8 + √x)/(8 - √x))^⅕ + 1 = (16/(8 - √x))^⅕
But ((8 + √x)/(8 - √x)) + 1 - 1 = (16/(8 - √x)) - 1
With (16/(8 - √x)) = t⁵, equation simplifies to (t⁵ - 1)^⅕ = t - 1
Raising equation to power 5 and using polynomial expansion
t⁵ - 1 = t⁵ - 5t⁴ + 10t³ - 10t² + 5t - 1
=> - 5t⁴ + 10t³ - 10t² + 5t = 0
=> t³ - 2t² + 2t - 1 = 0
=> (t - 1)(t² - t + 1) = 0
With t = 1 (Reject t² - t + 1 = 0)
(16/(8 - √x)) = (1)⁵
=> √x = -8
Or x = 64

[@RealQinnMalloryu4]

x^5+x^5+x^5+x^5+x^5x^5 ➖ x^5 ➖ x^5 ➖ x^5 ➖ x^5=x^25+8x+8x+8x+8x+8x8x ➖ 8x ➖ 8x ➖ 8x ➖ 8x=40x^5 {x^25+40x^5}= 40x^30 4^10x^3^10 4^2^5x^3^2^5 2^21^1x3^1^1 1^2x^3^1 2x^3 (x ➖ 3x+2).( x^5 )^2➖ (8x^5)^2= {x^25 ➖ 64x^25}=64x^1 8^8x^1^1 2^32^3x 1^12^1x 2^1x (x ➖ 2x+1). x^5+x^5x^5+x^5+x^5x^5 ➖ x^5 ➖ x^5 ➖ x^5 ➖ x^5=x^25 +16x+16x+16x+16x+1616x ➖ 16x ➖ 16x ➖ 16x ➖ 16x=80x^5 {x^25+80x^5}= 80x^30 8^10x^3^10 8^2^5x3^2^5 8^1^1x^3^1^1 2^3x^3^1 2^1x^3^1 2x^3 (x ➖ 3x+2).

[@SidneiMV]

8 + √x = u
8 - √x = v
x > 0
u + v = 16
uv = 64 - x
u⅕ + v⅕ = 16⅕
(u⅕ + v⅕)⁵ = 16
u + v + 5(uv)⅕(u⅗ + v⅗) + 10(uv)⅖(u⅕ + v⅕) = 16
(uv)⅕(u⅗ + v⅗) + 2(uv)⅖(u⅕ + v⅕) = 0
(uv)⅕(u⅕ + v⅕)(u⅖ + v⅖ + (uv)⅕) = 0
u⅕ + v⅕ = 16⅕
(uv)⅕(u⅖ + v⅖ + (uv)⅕) = 0
(uv)⅕ = 0 => uv = 0
64 - x = 0 => *x = 64* [ √x = 8 ]
u⅖ + v⅖ + (uv)⅕ = 0
(u⅕ + v⅕)² = 256⅕
u⅖ + v⅖ + 2(uv)⅕ = 256⅕
(uv)⅕ = 256⅕ => uv = 256
64 - x = 256 => x = -2⁶3 [ √x = 4i√3 ]
x < 0 => NOT VALID

[@SidneiMV]

8 + 4i√3 = 4(2 + i√3)
8 - 4i√3 = 4(2 - i√3)
4⅕(2 + i√3)⅕ + 4⅕(2 - i√3)⅕ = 16⅕
(2 + i√3)⅕ + (2 - i√3)⅕ = 4⅕
Yes/Not