Question 28 - If 𝛼 and 𝛽 are zeroes of a polynomial 6x2 -5x+1, then - CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard

part 2 - Question 28 - CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10
part 3 - Question 28 - CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10

 

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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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