Question

If one Zero of the polynomial 2x2

-5x-(2k+1) is twice the other. Find both the

zeroes of the polynomial and the value of k

Answer 1

let one zero is
$\alpha$
and other is
$2 \alpha \\ 2 {x}^{2} - 5x - (2k + 1) = 0 \\ \alpha + 2 \alpha = \frac{5}{2} \\ 3 \alpha = \frac{5}{2} \\ \alpha = \frac{5}{6} \\ other \: zero \: is \: 2 \times \frac{5}{6} = \frac{5}{3} \\ multiplication \: of \: zeros \\ \: \alpha \times 2\alpha = \frac{ - (2k + 1)}{2} \\ 2 { \alpha }^{2} = \frac{ - (2k + 1)}{2} \\ 2 \times \frac{25}{36} = \frac{ - (2k + 1)}{2} \\ 25 \times 2 = - 18(2k + 1) \\ 25 = - 9(2k + 1) \\ 25 = - 18k - 9 \\ - 18k = 25 + 9 \\ - 18k = 34 \\ k = - \frac{34}{18} \\ = \frac{ - 17}{9}$