Question

y²+3/2y-5 polynomials

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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

Answer 2

Step-by-step explanation:

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