Question

alpha and beeta are zeroes of the quadratic polynomial x^2-(k+6)x+2(2k-1). find the value of k if alpha+beeta=1/2alpha*beeta​

Answer 1

Answer:-

k = 7.

Explanation:-

Let , a = 1 ; b = -(k + 6) ; c = 2(2k - 1) = 4k - 2.

$we \: know \: that \\ \\ sum \: of \: the \: roots \: ( \alpha + \beta ) = \frac{ - b}{a} \\ \\ \alpha + \beta = \frac{ - ( - k - 6)}{1} \\ \\ product \: of \: the \: rooyts( \alpha \beta ) = \frac{c}{a} \\ \\ \alpha \beta = \frac{4k - 2}{1} \\ \\ \alpha \beta = 4k - 2 \\ \\ and \: also \: given \: that \: \alpha + \beta = \frac{1}{2} \times \alpha \beta \\ \\ k + 6 = \frac{1}{2} \times 4k - 2 \\ \\ k + 6 = \frac{2(2k - 1)}{2} \\ \\ k + 6 = 2k - 1 \\ \\ k - 2k = - 1 - 6 \\ \\ - k = - 7 \\ \\ k = 7$

Hence, the value of "k" is 7.