Question

If the sum of the zeros of the polynomial k y square + 2 y + 3 is equal to the product find the value of k

Answer 1

Let a and b are the zeros of quadratic equation ; Ky² + 2y -3 = 0
sum of zeros = -( coefficient of x)/coefficient of x²) = -(2)/K = -2/K
product of zeros = constant/coefficient of x²
= -3/K

A/C to question,
sum of zeros = 2 × product of zeros
-2/K = 2(-3)/K
-2 =-6. but -2≠ -6
here you can see that
LHS ≠ RHS

this question is given wrong data , it's not possible .

we can't find out K value by this condition .
but qudartic polynomial have real zeros then
Discrimant ≥ 0
( 2)² - 4(-3)K ≥ 0
1 + 3K ≥ 0
K ≥ -1/3
hence, for defining quadratic polynomial , K must be greater then -1/3

L

Answer 2

Step-by-step explanation:

Answer:

Given Polynomial : ky² - 2y - 3k

a = k

b = - 2

c = - 3k

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\textsf{Sum of Zeroes = Twice of Product of Zeroes}\\\\\\:\implies\sf \dfrac{-\:b}{a}=2 \times \dfrac{c}{a}\\\\\\:\implies\sf - \:b =2c\\\\\\:\implies\sf -( - 2) =2 \times ( - 3k)\\\\\\:\implies\sf 2 = 2 \times - 3k\\\\\\:\implies\sf \dfrac{2}{2} = -3k\\\\\\:\implies\sf1 = - 3k\\\\\\:\implies\sf \dfrac{1}{ - 3} = k\\\\\\:\implies\underline{\boxed{\sf k = \dfrac{ - \:1}{3} }}$