Find the sum of all real values of X satisfying the equation `(x^2-5x+5)^(x^2 + 4x -60) = 1` .

Options

A. 6
B. 5
C. 3
D. `-4`

Explanation

To solve the equation \((x^2 - 5x + 5)^{(x^2 + 4x - 60)} = 1\), we need to find all real values of \(x\) that satisfy this equation.

### Step 1: Identify conditions for the equation to equal 1
The expression \(a^b = 1\) can hold true under several conditions:
1. \(a = 1\) (where \(b\) can be any real number)
2. \(a = -1\) (where \(b\) must be an even integer)
3. \(b = 0\) (where \(a\) is not equal to 0)

### Step 2: Case 1: \(x^2 + 4x - 60 = 0\)
We start with the case where the exponent is zero:
\[ x^2 + 4x - 60 = 0 \]
Factoring the quadratic:
\[ x^2 + 10x - 6x - 60 = 0 \implies (x + 10)(x - 6) = 0 \]
Setting each factor to zero gives:
\[ x + 10 = 0 \implies x = -10 \]
\[ x - 6 = 0 \implies x = 6 \]
Thus, from Case 1, we have two solutions: \(x = -10\) and \(x = 6\).

### Step 3: Case 2: \(x^2 - 5x + 5 = 1\)
Next, we consider the case where the base equals 1:
\[ x^2 - 5x + 5 = 1 \]
Rearranging gives:
\[ x^2 - 5x + 4 = 0 \]
Factoring:
\[ x^2 - 4x - x + 4 = 0 \implies (x - 4)(x - 1) = 0 \]
Setting each factor to zero gives:
\[ x - 4 = 0 \implies x = 4 \]
\[ x - 1 = 0 \implies x = 1 \]
Thus, from Case 2, we have two solutions: \(x = 4\) and \(x = 1\).

### Step 4: Case 3: \(x^2 - 5x + 5 = -1\)
Now, we check the case where the base equals -1:
\[ x^2 - 5x + 5 = -1 \]
Rearranging gives:
\[ x^2 - 5x + 6 = 0 \]
Factoring:
\[ x^2 - 3x - 2x + 6 = 0 \implies (x - 2)(x - 3) = 0 \]
Setting each factor to zero gives:
\[ x - 2 = 0 \implies x = 2 \]
\[ x - 3 = 0 \implies x = 3 \]
However, we need to check if the exponent is even for \(x = 3\):
\[ x^2 + 4x - 60 \text{ at } x = 3 \implies 3^2 + 4(3) - 60 = 9 + 12 - 60 = -39 \text{ (odd)} \]
Thus, \(x = 3\) is not valid. We only accept \(x = 2\) from this case.

### Step 5: Collect all valid solutions
From all cases, the valid solutions are:
- From Case 1: \(x = -10, 6\)
- From Case 2: \(x = 1, 4\)
- From Case 3: \(x = 2\)

### Step 6: Sum of all real values of \(x\)
Now, we sum all the valid solutions:
\[ \text{Sum} = -10 + 6 + 1 + 4 + 2 = -10 + 13 = 3 \]

### Final Answer
The sum of all real values of \(x\) satisfying the equation is \(3\).

---