If `x^(2)-63x-64=0` and p and n are integers such that `p^(n)=x`, which of the following CANNOT be a value for p?

Options

A. `-8`
B. `-4`
C. `-1`
D. `4`

Answer

To solve the equation \(x^2 - 63x - 64 = 0\) and determine which value of \(p\) cannot be achieved when \(p^n = x\), we will follow these steps:

### Step 1: Solve the quadratic equation
We start with the equation:
\[ x^2 - 63x - 64 = 0 \]
We can factor this equation. We look for two numbers that multiply to \(-64\) and add up to \(-63\). The numbers are \(-64\) and \(1\).

Thus, we can factor the equation as:
\[ (x - 64)(x + 1) = 0 \]

### Step 2: Find the roots
Setting each factor to zero gives us:
\[ x - 64 = 0 \quad \Rightarrow \quad x = 64 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

So, the solutions for \(x\) are \(64\) and \(-1\).

### Step 3: Analyze the values of \(p\)
We know \(p^n = x\) where \(p\) and \(n\) are integers. We need to check which values of \(p\) can yield \(x = 64\) or \(x = -1\).

1. **For \(x = 64\)**:
- \(p = 4\): \(4^3 = 64\) (valid)
- \(p = 8\): \(8^2 = 64\) (valid)
- \(p = -8\): \((-8)^2 = 64\) (valid)
- \(p = -4\): \((-4)^n\) cannot equal \(64\) for any integer \(n\) since \((-4)^n\) is positive only for even \(n\) and negative for odd \(n\).

2. **For \(x = -1\)**:
- \(p = -1\): \((-1)^1 = -1\) (valid)
- \(p = 1\): \(1^n = -1\) is not possible for any integer \(n\).
- \(p = -1\): \((-1)^n = -1\) for odd \(n\) (valid).

### Step 4: Identify which \(p\) cannot be a value
From our analysis:
- \(p = 4\), \(p = 8\), and \(p = -8\) can yield \(64\).
- \(p = -1\) can yield \(-1\).
- \(p = -4\) cannot yield \(64\) and cannot yield \(-1\) for any integer \(n\).

Thus, the value of \(p\) that **cannot** be achieved is:
\[ \boxed{-4} \]