Analyze the equation:
\[ x(x^2 + 3x) + |x - 1| + 6|k - 2| = 0 \implies x^3 + 3x^2 + |x - 1| + C = 0. \]
Substitute \( C = 6|k - 2| \):
\[ x^3 + 3x^2 + |x - 1| + C = 0. \]
Consider cases for absolute value:
Case 1: \( x \geq 1 \):
\[ x^3 + 3x^2 + (x - 1) + C = 0 \implies x^3 + 3x^2 + x + (C - 1) = 0. \]
Case 2: \( x < 1 \):
\[ x^3 + 3x^2 + (-x + 1) + C = 0 \implies x^3 + 3x^2 - x + (C + 1) = 0. \]
Finding the number of real solutions: The cubic functions yield at most one real root due to their monotonically increasing nature.
Evaluate the function: Since the cubic functions are monotonic, we conclude: There is one real solution in each case.
Thus, the number of real solutions is: 1.
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: