Question

Given that α and β are the zeroes of the quadratic polynomial x2 - (k+6)x + 2(2k - 1). find k if α + β = 1 * αβ/2 .

Answer 1

We know that
alpha+beta= k+6 & alpha*beta=2(2k-1)
As alpha,beta is the two roots of above polynomial so,
ALPHA+Beta=1*(alpha*Beta)/2
or,k+6=1*(2*(2k-1))/2
or,k+6=2k-1
or,k=6+1=7 (ans)

Answer 2

Answer:

• The required value of k is 7.

Step-by-step explanation:

We have been given that α and β are zeroes of x² –(k + 6)x + 2(2k –1). We have to find the find value of k if 1/α + 1/β =1/2.

Sum of Zeros:

Sum of Zeros = -b/a  

Sum of Zeros( α + β)= k + 6

Product of Zeros:

Product of Zeros = c/a  

Product of Zeros(α + β) = 4k - 2

Value of 1/α + 1/β =1/2:

1/α + 1/β =1/2

α + β/αβ = ½

α + β = ½ αβ

K + 6 = ½ * 4k - 2  

2(k + 6) = 4k - 2

2k + 12 = 4k - 2  

2k - 4k = -2 - 12  

- 2k = - 14  

2k = 14

K = 14/2

K = 7

Therefore, the required value of k is 7.