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:

Question

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:

Accepted Answer

To solve this problem:

1. **Find $$ \alpha + \beta $$:**  
   From the quadratic equation $$ 2x^2 - 6x + k = 0 $$, we know:  
   $$ \alpha + \beta = -\frac{b}{a} = -\frac{-6}{2} = 3 $$.

2. **Find $$ \alpha\beta $$:**  
   Using the quadratic equation formula:  
   $$ \alpha\beta = \frac{c}{a} = \frac{k}{2} $$.

3. **Relate to new roots:**  
   The roots of the equation $$ x^2 + px + p = 0 $$ are given as $$ (\alpha + \beta) $$ and $$ (\alpha\beta) $$. Thus:  
   - Sum of roots:  
     $$
     (\alpha + \beta) + (\alpha\beta) = -p
     $$
     Substituting values:  
     $$
     3 + \frac{k}{2} = -p
     $$
     Rearrange:  
     $$
     p = -3 - \frac{k}{2}
     $$
   - Product of roots:  
     $$
     (\alpha + \beta)(\alpha\beta) = p
     $$
     Substituting values:  
     $$
     (3)\left(\frac{k}{2}\right) = p
     $$
     Rearrange:  
     $$
     p = \frac{3k}{2}
     $$

4. **Solve for $$ k - p $$:**  
   Equate both expressions for $$ p $$:  
   $$
   -3 - \frac{k}{2} = \frac{3k}{2}
   $$
   Simplify:  
   $$
   -3 = k + k/4
   $$
   Solve for $$ k $$:  
   $$
   k = 9/4
   $$
   Substitute into expressions for $$ p $$:  
   $$
   p = -9/4
   $$
   
5. **Calculate $$ 8(k - p) $$:**  
   Substitute values of $$ k $$ and $$ p $$:  
   $$
   8(k - p) = 8\left(\frac{9}{4} - (-\frac{9}{4})\right)
   $$
   Simplify:  
   $$
   8(k - p) = 8\left(\frac{18}{4}\right) = 6
   $$

Thus, the value of **$$ 8(k - p) = 6 $$**