A Nice Radical Equation | 95% Failed To Solve! 

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.)

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 }

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

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).

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

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

Substitute x-21=y

X= 21

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) = a^4+b^4+ab(a^2+b^2)+a^2hb^2=1. 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.