# Q.1 # Ako # Alfa, beta # su korijeni jednadžbe # X ^ 2-2x + 3 = 0 # dobiti jednadžbu čiji su korijeni # alfa ^ 3-3 alfa ^ 2 + 5 alfa -2 # i # P ^ 3-beta ^ 2 + P + 5 #?
Odgovor
danoj jednadžbi # X ^ 2-2x + 3 = 0 #
# => X = (2pmsqrt (2 ^ 2-4 * 1 * 3)) / 2 = 1pmsqrt2i #
pustiti # alpha = 1 + sqrt2i i beta = 1-sqrt2i #
Sada pusti
# gama = alfa ^ 3-3 alfa ^ 2 + 5 alfa -2 #
# => gama = alfa ^ 3-3 alfa ^ 2 + 3 alfa -1 + 2alpha-1 #
# => Y = (a-1) ^ 3 + alfa-1 + # alfa
# => Y = (sqrt2i) ^ 3 + sqrt2i + 1 + # sqrt2i
# => Y = -2sqrt2i + sqrt2i + 1 + 1 = sqrt2i #
I neka
# Delta = P ^ 3-beta ^ 2 + P + 5 #
# => = Delta p ^ 2 (beta-1) + P + 5 #
# => Delta = (1-sqrt2i) ^ 2 (-sqrt2i) + 1 sqrt2i + 5 #
# => Delta = (- 1-2sqrt2i) (- sqrt2i) + 1 sqrt2i + 5 #
# => Delta = sqrt2i-4 + 1 = 5 sqrt2i + 2 #
Dakle, kvadratna jednadžba ima korijene #gamma i delta # je
# X ^ 2- (y + delta) X + = 0 gammadelta #
# => X ^ 2- (1 + 2) + x 1 x 2 = 0 #
# => X ^ 2-3x + 2 = 0 #
# Q.2 # Ako je jedan korijen jednadžbe # X ^ 2 + bx + c = 0 # biti trg drugog, Dokaži to # B ^ 3 + a ^ 2c + ac ^ 2-3abc #
Neka jedan korijen bude #alfa# tada će drugi korijen biti # A ^ 2 #
Tako # A ^ 2 + a-b / a #
i
# ^ 3 a = C / a #
# => A ^ 3-1-c / a-1 #
# => (A-1), (a ^ 2 + 1) + alfa = 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) #
# => A = (c-a) / (a-b) = 1 + (c-b) / (a-b) #
Sada #alpha # kao jedan od korijena kvadratne jednadžbe # X ^ 2 + bx + c = 0 # možemo pisati
# Aalfa ^ 2 + balpha + c = 0 #
# => Aa ((c-b) / (a-b)) ^ 2 + b ((c-b) / (a-b)) + c = 0 #
# => (Ci-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 #
dokazao
Alternativa
# Aalfa ^ 2 + balpha + c = 0 #
# => Aalfa + b + c / a = 0 #
# => Aa (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 + 2 + ca ^ C ^ 2a = 3abc #
