If the sum of the roots of the equation $x^{2}-(k+6) x+2(2 k-1)=0$ is equal to half of their product, then $k=$
(a) 6
(b) 7
(c) 1
(d) 5
The given quadric equation is $x^{2}-(k+6) x+2(2 k-1)=0$, and roots are equal
Then find the value of k.
Let $\alpha$ and $\beta$ be two roots of given equation
And, $a=1, b=-(k+6)$ and,$c=2(2 k-1)$
Then, as we know that sum of the roots
$\alpha+\beta=\frac{-b}{a}$
$\alpha+\beta=\frac{-\{-(k+6)\}}{1}$
$=(k+6)$
And the product of the roots
$\alpha \cdot \beta=\frac{c}{a}$
$\alpha \beta=\frac{2(2 k-1)}{1}$
$=2(2 k-1)$
According to question, sum of the roots $=\frac{1}{2} \times$ product of the roots
$k+6=2 k-1$
$6+1=2 k-k$
$7=k$
Therefore, the value of $k=7$
Thus, the correct answer is
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