Find the value of k such that the polynomial x2 - (k + 6)x + 2(2k - 1) has the sum of its zeros equal to half of their product.
Answer: The value of k is 7.
We will use the coefficients to find the sum and the product of the zeros of the given polynomial.
The standard quadratic equation is ax2 + bx + c = 0.
Now, comparing the given equation with the standard form, we get
a = 1
b = -(k + 6)
c = 2(2k - 1)
Now, the sum of zeros of a polynomial = -b/a
= -(-(k + 6))
= k + 6
And, product of zeros of a polynomial = c/a
= 2(2k - 1)
= 4k - 2
Now, according to question,
Sum of zeros = 1/2 of product of zeros
k + 6 = 1/2 (4k - 2)
k + 6 = 2k - 1
2k - k = 6 + 1
k = 7
Therefore, the value of k for which the sum of the zeros of the polynomial is equal to half of their product is 7.