If one zero of the polynomial …

Let $\alpha$ and 2$\alpha$ are the zeroes of the polynomial 2x² - 5x - (2k + 1).
Then, 2$\alpha$² - 5$\alpha$ - (2k + 1) = 0
and 2 (2$\alpha$)² - 5(2$\alpha$) - (2k + 1) = 0
$\Rightarrow$ 2$\alpha$² - 5$\alpha$ = 2k + 1 ...(i)
and 8$\alpha^2$ - 10$\alpha$ = 2k + 1 ...(ii)
From Eqs. (i) and (ii), we get
2$\alpha^2$ - 5a = 8$\alpha^2$ - 10$\alpha$ $\Rightarrow$ 6$\alpha^2$ = 5$\alpha$ $\Rightarrow$ $\alpha=\frac{5}{6}$ $[\because \alpha \neq 0]$
$\therefore$ 2$\alpha$ = $\frac{5}{6} \times 2$ $=\frac{5}{3}$
Thus, the zeroes of the polynomial are $\frac{5}{6}$ and $\frac{5}{3}$
Now, substituting $\alpha=\frac{5}{6}$ in Eq. (i), we get 
2 $\times$ $\frac{25}{36}-\frac{25}{6}$ = 2k + 1
$\Rightarrow$ 2k + 1 = $\frac{50-150}{36}$ $\Rightarrow 2 k+1$ = $-\frac{100}{36}$
$\Rightarrow$ 2k = $-\frac{100}{36}-1$ $\Rightarrow$ $2 k=$ $-\frac{136}{36}$
$\Rightarrow$ k = - $\frac{68}{36}$ = $\frac{-17}{9}$