# Q.1 # Li # alfa, beta # jsou kořeny rovnice # x ^ 2-2x + 3 = 0 # získat rovnici, jejíž kořeny jsou # alfa ^ 3-3 alfa ^ 2 + 5 alfa -2 # a # beta ^ 3-beta ^ 2 + beta + 5 #?
Odpovědět
dané rovnice # x ^ 2-2x + 3 = 0 #
# => x = (2pmsqrt (2 ^ 2-4 * 1 * 3)) / 2 = 1pmsqrt2i #
Nechat # alpha = 1 + sqrt2i a beta = 1-sqrt2i #
Teď ať
# gamma = alfa ^ 3-3 alfa ^ 2 + 5 alfa -2 #
# => gamma = alfa ^ 3-3 alfa ^ 2 + 3 alfa -1 + 2alfa-1 #
# => gamma = (alfa-1) ^ 3 + alfa-1 + alfa #
# => gamma = (sqrt2i) ^ 3 + sqrt2i + 1 + sqrt2i #
# => gamma = -2sqrt2i + sqrt2i + 1 + sqrt2i = 1 #
A nechte
# delta = beta ^ 3-beta ^ 2 + beta + 5 #
# => delta = beta ^ 2 (beta-1) + beta + 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 rovnice s kořeny #gamma a delta # je
# x ^ 2- (gamma + delta) x + gammadelta = 0 #
# => x ^ 2- (1 + 2) x + 1 * 2 = 0 #
# => x ^ 2-3x + 2 = 0 #
# Q.2 # Jestliže jeden kořen rovnice # ax ^ 2 + bx + c = 0 # být čtvercem druhého, Dokázat to # b ^ 3 + a ^ 2c + ac ^ 2 = 3abc #
Nechť je jeden kořen # alpha # pak další kořen bude # alpha ^ 2 #
Tak # alpha ^ 2 + alpha = -b / a #
a
# alpha ^ 3 = c / a #
# => alpha ^ 3-1 = c / a-1 #
# => (alfa-1) (alfa ^ 2 + alfa + 1) = c / a-1 = (c-a) / a #
# => (alfa-1) (- b / a + 1) = (c-a) / a #
# => (alfa-1) ((a-b) / a) = (c-a) / a #
# => (alfa-1) = (c-a) / (a-b) #
# => alfa = (c-a) / (a-b) + 1 = (c-b) / (a-b) #
Nyní #alpha # je jedním z kořenů kvadratické rovnice # ax ^ 2 + bx + c = 0 # můžeme psát
# 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 #
se ukázala
Alternativní
# 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 #
