Question

The number of integers that satisfy the equality \((x^2-5x+7)^{x+1} = 1\) is

• A. 2
• B. 3
• C. 5
• D. 4

Answer 1

The Correct Option is B

For \( a^b = 1 \), the following cases hold:

• \( a = 1 \): valid for any \( b \)
• \( a = -1 \): valid only if \( b \) is even
• \( b = 0 \): valid for any \( a \neq 0 \)

Step-by-step Evaluation:

Case 1: \( x + 1 = 0 \Rightarrow x = -1 \)

Any non-zero base to the power 0 equals 1. Check the base: \[ (x^2 - 5x + 7) = (-1)^2 + 5 + 7 = 1 + 5 + 7 = 13 \neq 0 \] So, this is valid. ✅ \( x = -1 \) is a solution.

Case 2: \( x^2 - 5x + 7 = 1 \)

Solve: \[ x^2 - 5x + 6 = 0 \Rightarrow (x - 2)(x - 3) = 0 \Rightarrow x = 2, \; x = 3 \] Both are valid integer solutions. 
✅ \( x = 2 \) and \( x = 3 \) are solutions.

Case 3: \( x^2 - 5x + 7 = -1 \)

\[ x^2 - 5x + 8 = 0 \Rightarrow \text{Discriminant} = 25 - 32 = -7 < 0 \] No real roots. ❌ No integer solution.

Final Answer:

The total number of integer solutions is: \[ \boxed{3} \] Correct Option: (B)