Question

y2+3/2√5y-5 find zero

plz explain every step
like why I am doin this​

Answer 1

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

Answer 2

Answer:

hope it helps you

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