Question

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

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

= - 5

Step-by-step explanation: