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

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Solution

Let $P (x) = 3 x^{2} + 2 k x + x - k - 5$
$= 3 x^{2} + x (2 k + 1) - (k + 5)$
i.e; $a x^{2} + b x + c$
$a = 3; b = 2 k + 1; c = - (k + 5)$
1) Sum of zeroes = $\frac{- b}{a} = \frac{- (2 k + 1)}{3}$
2) Product of zeroes = $\frac{c}{a} = \frac{- (k + 5)}{3}$
According to given
$- (2 k + 1) / 3 = \frac{1}{2} [- (k + 5) / 3]$
$2 (2 k + 1) = k + 5$
$4 k + 2 = k + 5$
$3 k = 3 \Rightarrow k = 1$

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