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

Let $P\left(x\right)=3x^2+2kx+x-k-5$

$=3x^2+x(2k+1)-(k+5)$

i.e; $a{x}^2+bx+c$

$a=3;b=2k+1;c=-(k+5)$

1) Sum of zeroes = $\frac{-b}{a}=\frac{-(2k+1)}{3}$

2) Product of zeroes = $\frac{c}{a}=\frac{-(k+5)}{3}$

According to given

$-(2k+1)/3=\frac{1}{2}[-(k+5\left)/3\right]$

$2(2k+1)=k+5$

$4k+2=k+5$

$3k=3\Rightarrow k=1$