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 can analyze the cases where the expression can equal 1.

### Step 1: Identify Cases
The expression \((a)^b = 1\) can hold true under the following conditions:
1. \(a = 1\) (where \(b\) can be any real number)
2. \(a \neq 0\) and \(b = 0\) (where \(a\) can be any non-zero real number)
3. \(a = -1\) and \(b\) is an even integer (where \(a\) can be -1 and \(b\) must be even)

### Step 2: Case 1: \(x^2 - 5x + 5 = 1\)
Set \(x^2 - 5x + 5 = 1\):
\[ x^2 - 5x + 4 = 0 \]
Factoring gives:
\[ (x - 4)(x - 1) = 0 \]
Thus, \(x = 1\) or \(x = 4\).

### Step 3: Case 2: \(x^2 + 4x - 45 = 0\)
Set \(x^2 + 4x - 45 = 0\):
\[ x^2 + 4x - 45 = 0 \]
Factoring gives:
\[ (x + 9)(x - 5) = 0 \]
Thus, \(x = -9\) or \(x = 5\).

We need to check that \(x^2 - 5x + 5 \neq 0\) for these values:
- For \(x = -9\):
\[ (-9)^2 - 5(-9) + 5 = 81 + 45 + 5 = 131 \neq 0 \]
- For \(x = 5\):
\[ (5)^2 - 5(5) + 5 = 25 - 25 + 5 = 5 \neq 0 \]

Both values are valid.

### Step 4: Case 3: \(x^2 - 5x + 5 = -1\)
Set \(x^2 - 5x + 5 = -1\):
\[ x^2 - 5x + 6 = 0 \]
Factoring gives:
\[ (x - 2)(x - 3) = 0 \]
Thus, \(x = 2\) or \(x = 3\).

We need to check that \(x^2 + 4x - 45\) is even for these values:
- For \(x = 2\):
\[ (2)^2 + 4(2) - 45 = 4 + 8 - 45 = -33 \text{ (odd, rejected)} \]
- For \(x = 3\):
\[ (3)^2 + 4(3) - 45 = 9 + 12 - 45 = -24 \text{ (even, accepted)} \]

### Step 5: Collect All Valid Solutions
The valid solutions from all cases are:
- From Case 1: \(x = 1, 4\)
- From Case 2: \(x = -9, 5\)
- From Case 3: \(x = 3\)

Thus, the valid solutions are \(x = 1, 4, -9, 5, 3\).

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

### Final Answer
The sum of all real values of \(x\) satisfying the equation is \(\boxed{4}\).