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

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Hint: We are going to use the formula of sum of roots and product of roots to solve this question.

Let us consider $\alpha $ and $\beta $ to be the roots of the polynomial.
If we have a quadratic equation\[a{x^2} + bx + c = 0\],
Then, sum of roots, \[\alpha + \beta = - \dfrac{b}{a}\]
And, product of roots, \[\alpha \beta = \dfrac{c}{a}\]
Therefore, if we put the values in the above equations,
\[\alpha + \beta = (k + 6).....(1)\]
\[\alpha \beta = 4k - 2......(2)\]
According to question, the condition given is sum of zeroes is equal half their product,
Therefore,
\[\alpha + \beta = \dfrac{{\alpha \beta }}{2}\]

Note: Keep in mind that it is given that the sum of the zeroes is half the product and not the other way around. It is very important to frame the equation correctly otherwise the entire solution will become wrong.
From \[(1)\],\[(2)\] we get,
\[k + 6 = 2k - 1\]
On further solving, we get,
\[k = 7\]