Find the value of $k$ such that the quadratic equation $x^{2} - (k + 6) x + 2 (2 k - 1) = 0$ has sum of the roots as half of their product.

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Solution

The correct option is B 7
The given quadratic equation is
$x^{2} - (k + 6) x + 2 (2 k - 1) = 0$
Here, $a = 1, b = - (k + 6), c = 2 (2 k - 1)$
Sum of the roots $= \frac{- b}{c} = k + 6$
Product of the roots $= \frac{c}{a} = 2 (2 k - 1)$
According to the given condition :
Sum of the roots $= \frac{1}{2} \times$ Product of the roots
$k + 6 = \frac{1}{2} \times 2 (2 k - 1)$
$k + 6 = 2 k - 1$
$∴ k = 7$

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