Question

x^7 - 17x^5 + 16x^3 = 0 rozwiąż równanie

Answer 1

x^3(x^4-17x^2 +16)=0

x^3=0

x=0  (3-krotny)

 

x^4-17x^2+16=0

x^2=t

 

t^2 -17t+16=0

Δ=b^2-4ac=289-64=225    √Δ=15

t1=(-b-√Δ)/2a=(17-15)/2=2/2=1

t2=(-b+√Δ)/2a=(17+15)/2=32/2=16

x^2=1          ∨      x^2=16

x=±1                      x=±4

 

odp:

x∈{ -4 , -1 , 0 , 1 , 4}

Answer 2

x⁷ - 17x⁵ + 16x³ = 0  

x³ (x⁴ - 17x² + 16) = 0

x³ = 0

x = 0

lub

x⁴ - 17x² + 16 = 0

x²  = t

 

t² - 17t + 16 = 0

∆ = b² - 4ac

∆ = (-17)² - 4 * 1 * 16 = 289 – 64 = 225

√∆ = √225 = 15

t₁ = (-b - √∆)/2a = (17 – 15)/2 = 2/2 = 1

t₂ = (-b + √∆)/2a = (17 + 15)/2 = 32/2 = 16

 

x² = 1

x² - 1 = 0

(x-1)(x+1) = 0

x -1 = 0  lub  x + 1 = 0

x = 1  lub  x = -1

 

lub

x² = 16

x² - 16 = 0

(x-4)(x+4) = 0

x-4 = 0  lub  x+4 = 0

x = 4  lub  x = -4

 

rozwiązaniem rozwiązania jest :

x ϵ { 0 , 1 ,-1, 4, -4 }