Question

HEYY ❤❤❤❤


find the value of k such that the polynomial x^2-(k+6)x+2(2k-1) has sum of its zeros equal to half of their product​

Answer 1

Given..

-b/a=c/2a

=>(k+6)=2(2k-1)/2

=>k+6=2k-1

=>k=7

Therefore your answer is 7.

Answer 2

p(x) = x²-(k+6)+2(2k-1)

let the zeros are alpha and beta

we know. , alpha +beta = -b/a

Alpha ×beta = C/ A

according to question ,

alpha +beta =1/2× Alpha × beta

-b/a =1/2×c/a

- -(k+6)/1 = 1/2×2(2k-1)/1

(k+6) = (2k-1)

k-2k =-1-6

-k = -7

k =7

Hope it helps you gain something.