One zero of $2x^2-9x+k$ is twice the other zero. What is the value of $k$?

Let the zeroes be $m$ and 2$m$. Sum: $3m=-\frac{b}{a}=\frac{9}{2}$, so $m=\frac{3}{2}$. Product: $m\cdot 2m=2m^2=\frac{c}{a}=\frac{k}{2}$, so $k=4m^2=4\times \frac{9}{4}=9$. Check: $2x^2-9x+9=(2x-3)(x-3)$, zeroes $\frac{3}{2}$ and 3, and 3 is indeed twice $\frac{3}{2}.\frac{9}{2}$ stops at the product $2m^2$ and forgets that the product equals $\frac{k}{2}$, not $k$; 18 doubles once too often; 6 comes from taking $m=3$ as the smaller zero.