Question

The number of real solutions of the equation$x \left( x^2 + 3|x| + 5|x - 1| + 6|x - 2| \right) = 0$is ______.

Answer 1

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.