# # Q.1 ak # Alfa, beta # sú korene rovnice # X ^ 2-2x + 3 = 0 # získať rovnicu, ktorej korene sú # alfa ^ 3-3 alfa ^ 2 + 5 alfa -2 # a # P ^ 3-beta ^ 2 + p + 5 #?
odpoveď
daná rovnica # X ^ 2-2x + 3 = 0 #
# => X = (2pmsqrt (2 ^ 2-4 * 1 * 3)) / 2 = 1pmsqrt2i #
nechať # alpha = 1 + sqrt2i a beta = 1-sqrt2i #
Teraz nech
# gamma = alfa ^ 3-3 alfa ^ 2 + 5 alfa -2 #
# => gama = alfa ^ 3-3 a ^ 2 + 3 alfa-1 + 2alfa-1 #
# => Y = (a-1) ^ 3 + alfa-1 + alfa #
# => Y = (sqrt2i) ^ 3 + sqrt2i + 1 + sqrt2i #
# => Y = -2sqrt2i + sqrt2i + 1 + sqrt2i = 1 #
A nechajme
# Delta = p ^ 3-beta ^ 2 + p + 5 #
# => Delta = beta ^ 2 (beta-1) + p + 5 #
# => Delta = (1-sqrt2i) ^ 2 (-sqrt2i) + 1-sqrt2i + 5 #
# => Delta = (- 1-2sqrt2i) (- sqrt2i) + 1-sqrt2i + 5 #
# => Delta = sqrt2i-4 + 1-sqrt2i + 5 = 2 #
Takže kvadratická rovnica s koreňmi #gamma a delta # je
# X ^ 2- (gama + delta) x + gammadelta = 0 #
# => X ^ 2- (1 + 2) x + 1 * 2 = 0 #
# => X ^ 2-3x + 2 = 0 #
# # Q.2 Ak jeden koreň rovnice # Ax ^ 2 + bx + c = 0 # byť námestím druhého, Dokážte, že # B ^ 3 + a ^ 2c + ac ^ 2 = 3abc #
Nech je jeden koreň # Alfa # potom bude iný koreň # Alfa ^ 2 #
tak # Alfa ^ 2 + alfa = -b / a #
a
# Alfa ^ 3 = c / a #
# => Alfa ^ 3-1 = c / a-1 #
# => (A-1), (a-2 ^ + alfa + 1) = C / A-1 = (c-a) / a #
# => (A-1), (- B / A + 1) = (c-a) / a #
# => (A-1), ((A-B) / A) = (c-a) / a #
# => (A-1) = (c-a) / (a-b) #
# => Alfa = (c-a) / (A-B) + 1 = (c-b) / (a-b) #
teraz #alpha # je jedným z koreňov kvadratickej rovnice # Ax ^ 2 + bx + c = 0 # môžeme písať
# Aalpha ^ 2 + balpha + c = 0 #
# => A ((c-b) / (a-b)) ^ 2 + b ((c-b) / (a-b)) + c = 0 #
# => A (c-b) ^ 2 + b (c-b) (a-b) + c (a-b) ^ 2 = 0 #
# => Ac ^ 2-2abc + ab ^ 2 + ABC-ab ^ 2-b ^ 2c + b ^ 3 + ca ^ 2-2abc + b ^ 2c = 0 #
# => B ^ 3 + a ^ 2c + ac ^ 2 = 3abc #
ukázalo
alternatívne
# Aalpha ^ 2 + balpha + c = 0 #
# => Aalpha + B + C / alfa = 0 #
# => A (c / a) ^ (1/3) + b + c / ((C / A) ^ (1/3)) = 0 #
# => C ^ (1/3) a ^ (2/3) + c ^ (2/3) a ^ (1/3) = - b #
# => (C ^ (1/3) a ^ (2/3) + c ^ (2/3) a ^ (1/3)) ^ 3 = (- b) ^ 3 #
# => (C ^ (1/3) a ^ (2/3)) ^ 3+ (c ^ (2/3) a ^ (1/3)) ^ 3 + 3c ^ (1/3) a ^ (2/3) XXC ^ (2/3) a ^ (1/3) (c ^ (1/3) a ^ (2/3) + c ^ (2/3) a ^ (1/3)) = (- b) ^ 3 #
# => Ca ^ 2 + c ^ 2a + 3ca (-b) = (- b) ^ 3 #
# => B ^ 3 + ca ^ 2 + c ^ 2a = 3abc #
