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Ex 2.4, 3 If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b. For a cubic Polynomial p(x) = ax3 + bx2 + cx + d With zeroes Ξ±, 𝛽 and Ξ³ We have 𝛂 + 𝛽 + 𝛄 = (βˆ’π’ƒ)/𝒂 𝛂"𝛽" + 𝛽𝛄 + 𝛄𝛂 = 𝒄/𝒂 𝛂"𝛽" 𝛄= (βˆ’π’…)/𝒂 Now, p(x) = x3 βˆ’ 3x2 + x + 1 Comparing with p(x) = Ax3 + Bx2 + Cx + D, A = 1, B = βˆ’3, C = 1 and D = 1 Zeroes are 𝜢 = a βˆ’ b, 𝜷 = a and 𝜸 = a + b Sum of zeroes Sum of zeroes = (βˆ’π΅)/𝐴 𝜢 + 𝜷 + 𝜸 = (βˆ’π‘©)/𝑨 a βˆ’ b + a + a + b = 3 3a = 3 a = 1 Sum of Product zeroes Sum of Product zeroes = 𝐢/𝐴 𝜢𝜷+ 𝜷 𝜸 + 𝜸 𝜢 = π‘ͺ/𝑨 (a – b)a + a(a + b) + (a + b) (a βˆ’ b) = 1 a2 – ba + a2 + ab + a2 – b2 = 1 a2 + a2 + a2 βˆ’b2 = 1 3a2 βˆ’ b2 = 1 Putting a = 1 3(1)2 βˆ’ b2 = 1 3 βˆ’ b2 = 1 3 – 1 = b2 b2 = 2 b = Β± √𝟐 Thus, a = 1 and b = Β± √𝟐

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.