If the sum of the roots of the equation x2−(k+6)x+2(2k−1)=0


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




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

$k+6=2 k-1$

$6+1=2 k-k$


Therefore, the value of $k=7$

Thus, the correct answer is

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