For $5x^2-kx-2$, the sum of the zeroes is twice the product of the zeroes. Find $k$.

Sum $=-\frac{b}{a}=\frac{k}{5}$; product $=\frac{c}{a}=-\frac{2}{5}$. The condition gives $\frac{k}{5}=2\times \left(-\frac{2}{5}\right)=-\frac{4}{5}$, so $k=-4$. (Then $5x^2+4x-2$ has sum $-\frac{4}{5}$ and product $-\frac{2}{5}$, and $-\frac{4}{5}=2\times \left(-\frac{2}{5}\right)$.) 4 loses the minus sign of the product; −2 sets sum = product instead of sum = 2 × product; −10 multiplies −2 by the leading coefficient 5 instead of dividing.