The roots of the quadratic equation `x^(2)+px+8=0` are `alpha ` and `beta`. Obtain the values of p, if
(i) `alpha=beta^(2)`
(ii) `alpha-beta=2`.

Answer

To solve the problem step by step, we will address each condition separately.

### Given:
The quadratic equation is \( x^2 + px + 8 = 0 \) with roots \( \alpha \) and \( \beta \).

### Step 1: Use Vieta's Formulas
From Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta = -p \)
- The product of the roots \( \alpha \beta = 8 \)

### Step 2: Solve for \( p \) when \( \alpha = \beta^2 \)
1. Substitute \( \alpha = \beta^2 \) into the product of the roots:
\[ \beta^2 \cdot \beta = 8 \implies \beta^3 = 8 \]
Thus, solving for \( \beta \):
\[ \beta = 2 \]
2. Now substitute \( \beta \) back to find \( \alpha \):
\[ \alpha = \beta^2 = 2^2 = 4 \]
3. Now, use the sum of the roots to find \( p \):
\[ \alpha + \beta = 4 + 2 = 6 \implies -p = 6 \implies p = -6 \]

### Step 3: Solve for \( p \) when \( \alpha - \beta = 2 \)
1. From the condition \( \alpha - \beta = 2 \), we can express \( \alpha \) in terms of \( \beta \):
\[ \alpha = \beta + 2 \]
2. Substitute this into the product of the roots:
\[ (\beta + 2) \beta = 8 \implies \beta^2 + 2\beta - 8 = 0 \]
3. Solve the quadratic equation using the quadratic formula:
\[ \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2} \]
\[ = \frac{-2 \pm 6}{2} \]
Thus, the solutions for \( \beta \) are:
\[ \beta = 2 \quad \text{or} \quad \beta = -4 \]
4. For \( \beta = 2 \):
\[ \alpha = 2 + 2 = 4 \]
5. For \( \beta = -4 \):
\[ \alpha = -4 + 2 = -2 \]
6. Now calculate \( p \) for both cases:
- For \( \beta = 2 \) and \( \alpha = 4 \):
\[ \alpha + \beta = 4 + 2 = 6 \implies -p = 6 \implies p = -6 \]
- For \( \beta = -4 \) and \( \alpha = -2 \):
\[ \alpha + \beta = -2 - 4 = -6 \implies -p = -6 \implies p = 6 \]

### Final Values of \( p \):
Thus, the values of \( p \) are \( -6 \) and \( 6 \).

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