Question

Find the value of k such that the polynomial 2x2 + (k – 3)x - 3(2k + 1) has sum of its zeroes equal to twice
of their product. ​

Answer 1

Answer:

Step-by-step explanation:\

GIVEN

2x2 + (k – 3)x - 3(2k + 1)

2x2 + (k – 3)x - 6k-3

Let Alpha and beta be the zeroes of the polynomial

Alpha + beta= co efficient of X / co efficient of X^2

                     = (k-3)/2

alpha * Beta= constant/co efficient of X^2

                    =(-6k- 3)/2

ATQ,

(k-3)/2=2[(-6k- 3)/2]

(k-3)/2 = -(6k+3)         ( AS 2 GETS CANCELLED)

(k-3)=  -12k-6(2 is transfered to RHS)

(k-3)=  -12k-6

k+12k= -6+3

13k= -3

k= -3/13

HOPE THIS IS HELPFUL..........

THANK YOU!!!