Question

If the sum of the roots of the equation x2-(k+6)x+2(2k-1)=0is equal to one third of their products then k =

Answer 1

Step-by-step explanation:

${x}^{2} - (k + 6)x + 2(2k - 1) = 0 \\ on \: comparing \: with \: a {x}^{2} + bx + c = 0 \\ a = 1 \\ b = - (k + 6) \\ c = 2(2k - 1)$

Sum of zeroes=

$\alpha + \beta = \frac{ - b}{a} = \frac{- ( - (k + 6))}{1} = k + 6$

Product of zeroes=

$\alpha \beta = \frac{c}{a} = \frac{2(2k - 1)}{1} = 4k - 2$

Given

$sum \: of \: zeroes = \frac{1}{3} \times product \: of \: zeroes \\ k + 6 = \frac{4k - 2}{3} \\ 3(k + 6) = {4k - 2} \\ 3k + 18 = 4k - 2 \\ 4k - 3k = 18 + 2 \\ k = 20$