A Nice Radical Equation | 95% Failed To Solve!

A Nice Radical Equation | 95% Failed To Solve!

Welcome to another exciting algebra challenge! In this video, we solve a fascinating radical equation. Join us as we delve into this intriguing algebraic challenge and explore the elegant solutions. Whether you're a math enthusiast, a student preparing for competitions, or simply love solving problems, this video is for you. Let's solve this together and enhance our problem-solving skills!

Don't forget to like, subscribe, and hit the notification bell to stay updated with more Math Olympiad challenges and math problem-solving content!

🔍 In this video:

Detailed walkthrough of a challenging algebra problem.
Tips and tricks for solving complex algebraic equations.
Encouragement to enhance your problem-solving skills and mathematical thinking.

📣 Call to Action:

Have a go at the problem yourself before watching the solution!
Share your solutions and approaches in the comments below.
If you enjoyed this challenge, give it a thumbs up and subscribe for more intriguing math problems!

#algebra #radical #math #problemsolving #learnmaths #mathematics #matholympiadpreparation

🔗 Useful Links:

Don't forget to like, share, and subscribe for more Math Olympiad content!

Thank You for Watching!!

[潘博宇-kl]

X1=21, X2=21+(2)^(1/2)i, X3=21-(2)^(1/2)i.

[souzasilva]

Acho uma grande falta quando o professor não indica o conjunto em questão.(I think it's a big mistake when the teacher doesn't indicate the set in question.)

[허공]

let t = x - 22, (x - 20)^5 - 2 = (x - 22)^5,
=> (t + 2)^5 - t^5 - 2 = 10(t^4 + 4t^3 + 8t^2 + 8t + 3) = 10(t + 1)^2*(t^2 + 2t + 3) = 0
=> t = { -1, -1±√2i } => x = { 21, 21±√2i }

[raghvendrasingh]

Let y = x-21
(y+1)^5 - 2 = (y-1)^5
y^5^+5y^4+10y^3+10y^2+5y-1 = y^5 -5y^4 +10y^3-10y^2+5y-1
y^2 ( y^2+2) = 0
y = 0, y= i√2, y= - i√2
x=21, x= 21+i√2 , x= 21-i√2

[RealQinnMalloryu]

x^5;(x^5 ➖ 320) ➖ 2=( x^5 )^2➖ 315 ➖ 2={x^25 ➖ 315} ➖ 2=290 ➖(2)^2={290 ➖ 4}=286 10^20^2^43 10^2^102^43^1 2^5^2^2^5^2^1^1 1^1^1^1^1^2 1^2 (x ➖ 2x+1).(x)^2 ➖ (22)^2={x^2 ➖484}=482 10^40^2^42 10^40^2^6^7 10^4^10^2^6^7 2^5^4^2^5^2^6^7^1 1^1^4^1^1^1^3^21^1 2^2^3^2 1^13^2 3^2 (x ➖ 3x+2).

[Chacal]

x = a+20, a⁵-2 = y⁵=> a⁵-y⁵ = 2, y = x-22 = a-2=> a-y = 2 => a²+y² = 4 +2ay, (a-y)³ =a³-y³-3ay(a-y)
a³-y³ = 8+6ay => (a³-y³)(a²+y²) = (8+6ay)(4 +2ay)=> a⁵-y⁵-a²y²(y-a) = 12a²y²+40ay+32
10a²y²+40ay+30 = 0 = 10(a²y²+4ay+3) = 10 (ay+3)(ay+1), y = a-2 => a(a-2)+3 = 0 =(a-1)²+2 >0
or a(a-2)+1 = 0 => a = 1 => x = a+20 = 21

[RashmiRay-cy]

Let x-20=a and [(x-20)^5 -2]^1/5=b. Thus, a-b=2 and a^5-b^5=2 > (a^5-b^5)/(a-b) = Let ab=t. Given a-b=2, a^2+b^2=2(t+2) and a^4+b^4=2(t^2+8t+8). So, 2(t^2+8t+8) + 2t(t+2) + t^2=1 > 5t^2+20t+15=0 > (t+1)(t+3)=0 > t=-1, -3. t=ab=-3 > a^2-2a+3=0, with no real solutions. t=-1 > (a-1)^2=0 > a=x-20 = 1 > x=21.

[key_board_x]

⁵√[(x - 20)⁵ - 2] = x - 22
{⁵√[(x - 20)⁵ - 2]}⁵ = (x - 22)⁵
(x - 20)⁵ - 2 = (x - 20 - 2)⁵
(x - 20)⁵ - 2 = [(x - 20) - 2]⁵ → let: p = x - 20
p⁵ - 2 = [p - 2]⁵
p⁵ - 2 = (p - 2)².(p - 2)².(p - 2)
p⁵ - 2 = (p² - 4p + 4).(p² - 4p + 4).(p - 2)
p⁵ - 2 = (p⁴ - 4p³ + 4p² - 4p³ + 16p² - 16p + 4p² - 16p + 16).(p - 2)
p⁵ - 2 = (p⁴ - 8p³ + 24p² - 32p + 16).(p - 2)
p⁵ - 2 = p⁵ - 8p⁴ + 24p³ - 32p² + 16p - 2p⁴ + 16p³ - 48p² + 64p - 32
- 2 = - 10p⁴ + 40p³ - 80p² + 80p - 32
- 10p⁴ + 40p³ - 80p² + 80p - 30 = 0
p⁴ - 4p³ + 8p² - 8p + 3 = 0 → the aim, if we are to continue effectively, is to eliminate terms to the 3rd power
p⁴ - 4p³ + 8p² - 8p + 3 = 0 → let: p = z - (b/4a) → where:
b is the coefficient for p³, in our case: - 4
a is the coefficient for p⁴, in our case: 1
p⁴ - 4p³ + 8p² - 8p + 3 = 0 → let: p = z - (- 4/4) → p = z + 1

(z + 1)⁴ - 4.(z + 1)³ + 8.(z + 1)² - 8.(z + 1) + 3 = 0
(z + 1)².(z + 1)² - 4.(z + 1)².(z + 1) + 8.(z² + 2z + 1) - 8z - 8 + 3 = 0
(z² + 2z + 1).(z² + 2z + 1) - 4.(z² + 2z + 1).(z + 1) + 8z² + 16z + 8 - 8z - 8 + 3 = 0
(z⁴ + 2z³ + z² + 2z³ + 4z² + 2z + z² + 2z + 1) - 4.(z³ + z² + 2z² + 2z + z + 1) + 8z² + 16z + 8 - 8z - 8 + 3 = 0
(z⁴ + 4z³ + 6z² + 4z + 1) - 4.(z³ + 3z² + 3z + 1) + 8z² + 8z + 3 = 0
z⁴ + 4z³ + 6z² + 4z + 1 - 4z³ - 12z² - 12z - 4 + 8z² + 8z + 3 = 0
z⁴ + 2z² = 0
z².(z² + 2) = 0

First case: z² = 0
z = 0 → recall: p = z + 1
p = 1 → recall: p = x - 20
x = p + 20
→ x = 21

Second case: (z² + 2) = 0
z² + 2 = 0
z² = - 2
z² = 2i²
z = ± i√2 → recall: p = z + 1
p = 1 ± i√2 → recall: p = x - 20
x = p + 20
x = 1 ± i√2 + 20
→ x = 21 ± i√2