Answer: By splitting the middle term y2 + 3√5/2y – 5 = 0 2y2 + 3√5y – 10 = 0 2y2 + (4√5y – √5y) – 10 = 0 2y2 + (4√5y – √5y) – 10 = 0 2y(y + 2√5) – √5(y + 2√5) = 0 (y + 2√5)(2y – √5) = 0 ⇒ y = – 2√5, √5/2 Verification: Sum of the zeroes = – (coefficient of x) ÷ coefficient of x2 α + β = – b/a (- 2√5) + (√5/2) = – (3√5)/2 = – 3√5/2 = – 3√5/2 Product of the zeroes = constant term ÷ coefficient of x2 α β = c/a (- 2√5)(√5/2) = – 5 – 5 = – 5Read more on Sarthaks.com – https://www.sarthaks.com/878358/find-zeroes-polynomial-25y-verify-relation-between-the-coefficients-zeroes-polynomial?show=878360#a878360 Reply
Answer:
By splitting the middle term y2 + 3√5/2y – 5 = 0 2y2 + 3√5y – 10 = 0 2y2 + (4√5y – √5y) – 10 = 0 2y2 + (4√5y – √5y) – 10 = 0 2y(y + 2√5) – √5(y + 2√5) = 0 (y + 2√5)(2y – √5) = 0 ⇒ y = – 2√5, √5/2 Verification: Sum of the zeroes = – (coefficient of x) ÷ coefficient of x2 α + β = – b/a (- 2√5) + (√5/2) = – (3√5)/2 = – 3√5/2 = – 3√5/2 Product of the zeroes = constant term ÷ coefficient of x2 α β = c/a (- 2√5)(√5/2) = – 5 – 5 = – 5Read more on Sarthaks.com – https://www.sarthaks.com/878358/find-zeroes-polynomial-25y-verify-relation-between-the-coefficients-zeroes-polynomial?show=878360#a878360
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