Roots of Quadratic Equations

For $2x^2 - 6x + k = 0$, the sum and product of roots are: $\alpha + \beta = 3, \quad \alpha\beta = \frac{k}{2}$

The roots of $x^2 + px + p = 0$ are $3$ and $\frac{k}{2}$. Sum of roots: $3 + \frac{k}{2} = -p \implies \frac{k}{2} + p = -3$ Product of roots: $3 \cdot \frac{k}{2} = p \implies \frac{k}{2} = \frac{p}{3}$

Substitute $\frac{k}{2} = \frac{p}{3}$ into the sum equation: $\frac{p}{3} + p = -3 \implies \frac{4p}{3} = -3 \implies p = -\frac{9}{4}$ Then $k = 2 \cdot \frac{p}{3} = 2 \cdot \left(-\frac{3}{4}\right) = -\frac{3}{2}$.

Now evaluate $8(k - p)$: $k - p = -\frac{3}{2} - \left(-\frac{9}{4}\right) = \frac{3}{4}$ $8(k - p) = 8 \times \frac{3}{4} = 6$