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Find the value of k such that the polynomial $x^{2}-(k+6)x+2(2k-1)$ has sum of its zeros equal to half of their product.
Given: Polynomial $x^{2}-(k+6)x+2(2k-1)$ has sum of its zeros equal to half of their product.
To do: To Find the value of $k$.
Solution:
Given Polynomial $x^{2}-(k+6)x+2(2k-1)$ is a quadratic polynomial.
On comparing it to 4ax^{2}+bx+c$ we have,
$a=1, b=-(k+6)$ and $2(2k-1)$
Let $\alpha$ and $\beta$ are the zeroes of the the given polynomial.
As known,
Sum of the zeroes, $\alpha+\beta=-b$
$\Rightarrow \alpha+\beta=-(-(k+6))=k+6$
Half of the Product of the zeroes, $\frac{\alpha.\beta}{2}=\frac{1}{2}×\frac{c}{a}=\frac{2(2k-1)}{2}=2k-1$
As given: $\alpha+\beta=\frac{\alpha.\beta}{2}$
$k+6=2k-1$
$\Rightarrow 2k-k=6+1$
$\Rightarrow k=7$
Therefore the value of $k=7$.
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