For the polynomial $x^2+10x+k$, the zeroes α and β satisfy 1/α + 1/β = −5. Find $k$.

$\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{-10}{k}$. Setting $\frac{-10}{k}=-5$ gives $k=2$. (Then $x^2+10x+2$ has sum −10 and product 2, and $\frac{-10}{2}=-5$.) −2 loses a sign in the division; 50 multiplies −10 by −5 instead of dividing; $\frac{1}{2}$ inverts the final step.