How can I determine the value of k for the polynomial x² + kx + 6x² + 2kx + 1 such that the sum of its zeroes is equal to half of their product?

To find the value of k such that the polynomial x² + kx + 6x² + 2kx + 1 has the sum of its zeroes equal to half of their product, we start by rewriting the polynomial in its standard form.

First, combine like terms:

x² + 6x² + kx + 2kx + 1 = (7x²) + (k + 2k)x + 1 = 7x² + (3k)x + 1

Now, we can denote the polynomial as:

f(x) = 7x² + 3kx + 1

According to Vieta’s formulas for a quadratic polynomial ax² + bx + c, the sum of the roots (α + β) is given by:

-(b/a) = -(3k/7)

And the product of the roots (αβ) is given by:

c/a = (1/7)

According to the problem, we are looking for the case where the sum of the roots equals half of their product:

-(3k/7) = 1/2 * (1/7)

Now, simplifying this equation:

-(3k/7) = 1/14

Multiplying both sides by -7 gives:

3k = -1/2

Now, dividing both sides by 3, we find:

k = -1/6

Thus, the value of k such that the polynomial has the sum of its zeroes equal to half of their product is -1/6.