find the zero of y2+3/2×√5y-5 by factorization method and verify the relation between between the zeros and coefficient of polynomial.
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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
= - 5
Step-by-step explanation:
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