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

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....

Answer 2

the value of k is 1...