To solve the equation (x - 1)^2 = |3x - 5|, we need to consider two cases: when the expression inside the absolute value is positive and when it is negative. Case 1: When 3x - 5 ≥ 0 In this case, the absolute value simplifies to 3x - 5. We can rewrite the equation as (x - 1)^2 = 3x - 5. Expanding the left side, we get x^2 - 2x + 1 = 3x - 5. Rearranging the terms, we have x^2 - 5x + 6 = 0. Factoring the quadratic equation, we get (x - 2)(x - 3) = 0. This equation has two solutions: x = 2 and x = 3. Case 2: When 3x - 5 < 0 In this case, the absolute value simplifies to -(3x - 5), which is equivalent to -3x + 5. We can rewrite the equation as (x - 1)^2 = -3x + 5. Expanding the left side, we get x^2 - 2x + 1 = -3x + 5. Rearranging the terms, we have x^2 + x - 4 = 0. Using the quadratic formula, we find that the solutions for this equation are x ≈ 1.561 and x ≈ -2.561. Overall, we have four solutions for the equation (x - 1)^2 = |3x - 5|: x = 2, x = 3, x ≈ 1.561, and x ≈ -2.561. To find the minimum solution denoted by a, we can observe that x ≈ -2.561 is the smallest among the four solutions. Thus, a ≈ -2.561. To find the integer m satisfying m - 1 < a < m, we can see that m = -3 satisfies -4 < -2.561 < -3. Therefore, the integer m satisfying the given condition is -3.