# Find the value of k such that the polynomial x^{2 }- (k + 6)x + 2(2k - 1) has sum of its zeroes equal to half of their product.

A polynomial is a type of expression in which the exponents of all variables should be a whole number.

## Answer: The value of k for which the sum of its zeroes equal to half of their product is 7.

We will use the coefficients to find the sum and product of zeros.

**Explanation:**

The standard quadratic equation is ax^{2} + bx + c = 0.

Now, comparing the given equation with the standard form, we get

a = 1

b = -(k + 6)

c = 2(2k - 1)

Now, sum of zero = -b/a

= -(-(k + 6))

= k + 6

And, product of zero = c/a

= 2(2k - 1)

= 4k - 2

Now, according to question,

Sum of zero = 1/2 of product of zeros

k + 6 = 1/2 (4k - 2)

k + 6 = 2k - 1

2k - k = 6 + 1

k = 7