Question

Α and β are zeroes of the quadratic polynomial x2-(k+6)x+2(2k-1). find the value of k if α + β = ½ αβ.

Answer 1

x2-(k+6)x+2(2k-1)

α+β=-b/a
=-(-(k+6))/1
=k+6

αβ=c/a
=2(2k-1)/1
=2(2k-1)

Now,
α + β = ½ αβ

k+6=1/2x2(2k-1)
k+6=2k-1
k=7

Hope it helps!!!

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.