The sum of all real values of x satisfying the equation `(x^(2) -5x+5)^(x^(2)+4x-45)=1` is :

Options

A. 8
B. 6
C. 5
D. 4

Answer

To solve the equation \((x^2 - 5x + 5)^{(x^2 + 4x - 45)} = 1\), we need to analyze the cases where an expression raised to a power equals 1.

### Step 1: Identify cases where \(a^b = 1\)
The equation \(a^b = 1\) can be satisfied under the following conditions:
1. \(a = 1\)
2. \(a = -1\) and \(b\) is even
3. \(b = 0\) and \(a \neq 0\)

Let \(a = x^2 - 5x + 5\) and \(b = x^2 + 4x - 45\).

### Step 2: Case 1 - \(b = 0\)
Set \(b = 0\):
\[ x^2 + 4x - 45 = 0 \]
Factoring the quadratic:
\[ (x + 9)(x - 5) = 0 \]
Thus, we have:
\[ x = -9 \quad \text{or} \quad x = 5 \]

### Step 3: Case 2 - \(a = 1\)
Set \(a = 1\):
\[ x^2 - 5x + 5 = 1 \]
This simplifies to:
\[ x^2 - 5x + 4 = 0 \]
Factoring gives:
\[ (x - 1)(x - 4) = 0 \]
Thus, we have:
\[ x = 1 \quad \text{or} \quad x = 4 \]

### Step 4: Case 3 - \(a = -1\) and \(b\) is even
Set \(a = -1\):
\[ x^2 - 5x + 5 = -1 \]
This simplifies to:
\[ x^2 - 5x + 6 = 0 \]
Factoring gives:
\[ (x - 2)(x - 3) = 0 \]
Thus, we have:
\[ x = 2 \quad \text{or} \quad x = 3 \]
Next, we need to check if \(b\) is even for these values.

- For \(x = 2\):
\[ b = 2^2 + 4(2) - 45 = 4 + 8 - 45 = -33 \quad (\text{not even}) \]
- For \(x = 3\):
\[ b = 3^2 + 4(3) - 45 = 9 + 12 - 45 = -24 \quad (\text{even}) \]
Thus, \(x = 3\) is valid.

### Step 5: Collect all valid solutions
The valid solutions we found are:
- From Case 1: \(x = -9, 5\)
- From Case 2: \(x = 1, 4\)
- From Case 3: \(x = 3\)

### Step 6: Calculate the sum of all solutions
Now, we sum all valid solutions:
\[ -9 + 5 + 1 + 4 + 3 = -9 + 5 + 8 = -4 + 3 = -1 \]

### Final Answer
The sum of all real values of \(x\) satisfying the equation is \(-1\).