Question

find the value of k such that 3x²+2kx+x-k-5 has the sum of zeroes as half of their product

Answer 1

hope it is helpful for you

Answer 2

Answer:

Step-by-step explanation:Let α  and  β are  zeroes  of  the  polynomial p(x) =3x²  + 2kx  +  x- k–  5

3x²  +  (2kx  +  x) - k–  5

3x²  + x (2k +  1) - k–  5

On comparing with ax²+bx+c=0

a= 3, b= 2k+1, c = -k-5

Sum of zeroes (α+β)= -b/a = -(2k+1)/3

α+β= -(2k+1)/3

Product of zeros(α.β)= c/a = -k-5/3

α.β= c/a = -k-5/3

Sum of zeroes (α+β) = ½ (Product of zeros(α.β))    [ GIVEN]

-(2k+1)/3 = ½ (-k-5/3)

-2k-1= ½ (-k-5)

2(-2k-1) = -k -5

-4k-2 = -k-5

-4k +k = -5 +2

-3k = -3

k = 3/3

k = 1

Hence, the value of k is 1.

HOPE THIS WILL HELP YOU....