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# Find the value of k such that the polynomial x^{2 }- (k + 6)x + 2(2k - 1) has the sum of its zeros 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 is 7.

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

**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, 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.

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