If one zero of the polynomial …

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