Last updated at Aug. 2, 2021 by Teachoo
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Ex 2.4, 3 If the zeroes of the polynomial x3 β 3x2 + x + 1 are a β b, a, a + b, find a and b. For a cubic Polynomial p(x) = ax3 + bx2 + cx + d With zeroes Ξ±, π½ and Ξ³ We have π + π½ + π = (βπ)/π π"π½" + π½π + ππ = π/π π"π½" π= (βπ )/π Now, p(x) = x3 β 3x2 + x + 1 Comparing with p(x) = Ax3 + Bx2 + Cx + D, A = 1, B = β3, C = 1 and D = 1 Zeroes are πΆ = a β b, π· = a and πΈ = a + b Sum of zeroes Sum of zeroes = (βπ΅)/π΄ πΆ + π· + πΈ = (βπ©)/π¨ a β b + a + a + b = 3 3a = 3 a = 1 Sum of Product zeroes Sum of Product zeroes = πΆ/π΄ πΆπ·+ π· πΈ + πΈ πΆ = πͺ/π¨ (a β b)a + a(a + b) + (a + b) (a β b) = 1 a2 β ba + a2 + ab + a2 β b2 = 1 a2 + a2 + a2 βb2 = 1 3a2 β b2 = 1 Putting a = 1 3(1)2 β b2 = 1 3 β b2 = 1 3 β 1 = b2 b2 = 2 b = Β± βπ Thus, a = 1 and b = Β± βπ
Ex 2.4 (Optional)
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