Find the value of k such that $3 x^{2} + 2 k x + x - k - 5$ has the sum of the zeros as half of their product.

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Solution

$3 x^{2} + 2 k x + x - k - 5 = 0$
$3 x^{2} + (2 k + 1) x + (- k - 5) = 0$
$a = 3; b = (2 k + 1); c = (- k - 5)$
Product of roots = $\frac{c}{a}$ ; sum of roots = $- \frac{b}{a}$
Sum= $- \frac{b}{a} = - \frac{(2 k + 1)}{3}$
Product = $\frac{c}{a} = \frac{(- k - 5)}{3}$
Sum = $\frac{1}{2}$ of product
$- \frac{(2 k + 1)}{3} = (\frac{1}{2}) \frac{(- k - 5)}{3}$
Or, $- 4 k - 2 = - k - 5$
Or, $- 3 k = - 3$
Or, $k = 1$

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