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Question 28 If 𝛼 and 𝛽 are zeroes of a polynomial 6π‘₯2 βˆ’ 5x + 1 then form a quadratic polynomial whose zeroes are 𝛼2 and 𝛽2 .Let p(x) = 6π‘₯2 βˆ’ 5x + 1 Roots of p(x) are 𝛼 and 𝛽 So, Sum of roots = 𝛼 + 𝛽 = (βˆ’(βˆ’5))/6 = πŸ“/πŸ” And, Product of roots = 𝛼𝛽 = 𝟏/πŸ” We need to find quadratic polynomial whose zeroes are 𝛼2 and 𝛽2 . So, Required polynomial = q(x) = x2 βˆ’ (Sum of Zeroes)x + Product of Zeroes = x2 βˆ’ (𝛼2 + 𝛽2)x + 𝛼2 Γ— 𝛽2 Using (a2 + b2) = (a + b)2 βˆ’ 2an = x2 βˆ’ [(𝛼 + 𝛽)2 βˆ’ 2𝛼𝛽] x + (𝛼𝛽)2 Putting 𝛼 + 𝛽 = 5/6 & 𝛼𝛽 = 1/6 = π‘₯^2βˆ’[(5/6)^2βˆ’2 Γ—1/6]π‘₯+(1/6)^2 = π‘₯^2βˆ’[25/36βˆ’1/3]π‘₯+1/36 = π‘₯^2βˆ’[(25 βˆ’ 12)/36]π‘₯+1/36 = 𝒙^πŸβˆ’[πŸπŸ‘/πŸ‘πŸ”]𝒙+𝟏/πŸ‘πŸ” We can multiply by 36 to make our equation cleaner = 36(π‘₯^2βˆ’[13/36]π‘₯+1/36) = πŸ‘πŸ”π’™^πŸβˆ’πŸπŸ‘π’™+𝟏

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo