Question

6. If sum of zeros of the polynomial ky^2+2y-3k is equal to twice their product.
Find the value of k.​

Answer 1

Answer:

$\boxed{k=1/3}$

Step-by-step explanation:

$\boxed{given}$

sum of the zeores of the polynomial

$\underline{ky^2+2y-3k}$

is equal to twicw thier products

$\boxed{Tofind}$

value of k

$\boxed{Answer}$

$\alpha + \beta = \frac{ - b}{a}$

-b=-(2)=-2

a=k

so we get

$\alpha + \beta = \frac{ - 2}{k}$

now product of the zeores

$\alpha \beta = \frac{c}{a}$

where

c=-3k

a=k

so we get

$\alpha \beta = \frac{ - 3k}{k}$

now according to the quesion

$\alpha + \beta = 2( \alpha \beta )$

inserting the values we get

$\frac{ - 2}{k} = 2( \frac{ - 3k}{k} )$

$\frac{ - 2}{k} = 2( - 3)$

$\frac{ - 2}{k} = - 6$

$- 2 = - 6k$

$\frac{ - 2}{ - 6} = k$

$\frac{1}{3} = k$

Answer 2

Your answer lies in the attachment.