Let $\alpha$ and $\beta$ be the two distinct roots of the equation $2x^2 - 6x + k = 0$, such that $(\alpha + \beta)$ and $\alpha\beta$ are the distinct roots of the equation $x^2 + px + p = 0$. Then, the value of $8 (k - p)$ is

For any quadratic equation $ax^2 + bx + c = 0$, we have:

Sum of roots = $-\tfrac{b}{a}$

Product of roots = $\tfrac{c}{a}$

If $\alpha$ and $\beta$ are roots, then $(x - \alpha)(x - \beta) = x^2 - (\alpha + \beta)x + \alpha\beta = 0$. Comparing with $ax^2 + bx + c = 0$, we get these relationships.

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For our first equation $2x^2 - 6x + k = 0$:

$a = 2$, $b = -6$, $c = k$

Sum of roots: $\alpha + \beta = -\frac{(-6)}{2} = \frac{6}{2} = 3$

Product of roots: $\alpha\beta = \frac{k}{2}$

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The second equation $x^2 + px + p = 0$ has roots $(\alpha + \beta)$ and $\alpha\beta$.

We already know:

$\alpha + \beta = 3$

$\alpha\beta = \frac{k}{2}$

So the roots of the second equation are $3$ and $\frac{k}{2}$.

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For the second equation $x^2 + px + p = 0$:

Sum of roots: $3 + \frac{k}{2} = -p$ ... (equation 1)

Product of roots: $3 \times \frac{k}{2} = p$ ... (equation 2)

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From equation 2: $p = \frac{3k}{2}$

Substituting into equation 1:

$3 + \frac{k}{2} = -\frac{3k}{2}$

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$3 = -\frac{3k}{2} - \frac{k}{2}$

$3 = -\frac{4k}{2}$

$3 = -2k$

$k = -\frac{3}{2}$

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$p = \frac{3k}{2} = \frac{3(-\frac{3}{2})}{2} = \frac{-\frac{9}{2}}{2} = -\frac{9}{4}$

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$8(k - p) = 8\left(-\frac{3}{2} - \left(-\frac{9}{4}\right)\right)$

$= 8\left(-\frac{3}{2} + \frac{9}{4}\right)$

Converting to common denominator: $-\frac{3}{2} = -\frac{6}{4}$

$= 8\left(-\frac{6}{4} + \frac{9}{4}\right)$

$= 8\left(\frac{3}{4}\right)$

$= 6$

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When dealing with problems involving roots as coefficients of other equations, we always use the sum and product of roots formulas. This creates a system of equations that can be solved systematically.

Answer: $8(k - p) = 6$