# Q.1 # Če # alpha, beta # so korenine enačbe # x ^ 2-2x + 3 = 0 # dobite enačbo, katere korenine so # alfa ^ 3-3 alfa ^ 2 + 5 alfa -2 in # beta ^ 3-beta ^ 2 + beta + 5 #?
Odgovor
dani enačbi # x ^ 2-2x + 3 = 0 #
# => x = (2pmsqrt (2 ^ 2-4 * 1 * 3)) / 2 = 1pmsqrt2i #
Let # alpha = 1 + sqrt2i in beta = 1-sqrt2i #
Zdaj pa pustite
# gama = alfa ^ 3-3 alfa ^ 2 + 5 alfa -2
# => gama = alfa ^ 3-3 alfa ^ 2 + 3 alfa -1 + 2alpha-1 #
# => gama = (alfa-1) ^ 3 + alfa-1 + alfa #
# => gamma = (sqrt2i) ^ 3 + sqrt2i + 1 + sqrt2i #
# => gamma = -2sqrt2i + sqrt2i + 1 + sqrt2i = 1 #
In pustite
# 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 #
Torej kvadratna enačba ima korenine #gamma in delta # je
# x ^ 2- (gama + delta) x + gammadelta = 0 #
# => x ^ 2- (1 + 2) x + 1 * 2 = 0 #
# => x ^ 2-3x + 2 = 0 #
# Q.2 # Če je en koren enačbe # ax ^ 2 + bx + c = 0 # biti kvadrat drugega, Dokaži, da # b ^ 3 + a ^ 2c + ac ^ 2 = 3abc #
Naj bo en koren # alfa # potem bo druga korenina # alfa ^ 2 #
Torej # alfa ^ 2 + alfa = -b / a #
in
# alpha ^ 3 = c / a #
# => alfa ^ 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) #
Zdaj #alpha # je eden od korenin kvadratne enačbe # ax ^ 2 + bx + c = 0 # lahko pišemo
# 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 #
dokazano
Alternativa
# aalpha ^ 2 + balpha + c = 0 #
# => aalpha + b + c / alpha = 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 #
