Find the value of k such that the polynomial x² + kx + 6x² + 2k + 1 has sum of its zeroes equal to half of their product

To solve for the value of k in the polynomial x² + kx + 6x² + 2k + 1, we first need to simplify the polynomial. Combining like terms, we get:

7x² + kx + 2k + 1 = 0

The sum of the roots (zeroes) of a quadratic equation ax² + bx + c = 0 is given by -b/a, and the product of the roots is given by c/a. Here, a = 7, b = k, and c = 2k + 1.

Thus, the sum of the roots is:

Sum = -k/7

And the product of the roots is:

Product = (2k + 1)/7

According to the problem, the sum of the roots should be equal to half of their product. This gives us the equation:

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

Multiplying both sides by 7 to eliminate the denominator results in:

-k = 1/2 * (2k + 1)

Multiplying through by 2 to clear the fraction:

-2k = 2k + 1

Rearranging gives:

-2k – 2k = 1

-4k = 1

So, dividing both sides by -4 leads us to:

k = -1/4

In conclusion, the value of k such that the sum of the zeroes of the polynomial equals half of their product is -1/4.

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