Question

If \( x |x - 3| + 3 |x - 2| + 1 = 0 \), then the number of real solutions is.

Hint:

When solving absolute value equations, consider the sign change at the points where the absolute value expressions are zero. Divide the solution into intervals and solve for each interval.

Answer 1

Consider the equation: \[x |x - 3| + 3 |x - 2| + 1 = 0\]To solve for \( x \), we analyze three intervals based on the critical points \( x = 2 \) and \( x = 3 \), where the absolute value expressions change sign. These intervals are \( x<2 \), \( 2 \leq x<3 \), and \( x \geq 3 \).Case 1: \( x<2 \)Here, \( |x - 3| = 3 - x \) and \( |x - 2| = 2 - x \). The equation simplifies to:\[x (3 - x) + 3 (2 - x) + 1 = 0\]\[3x - x^2 + 6 - 3x + 1 = 0\]\[-x^2 + 7 = 0\]\[x^2 = 7\]\[x = \pm \sqrt{7}\]Given \( x<2 \), the only valid solution is \( x = -\sqrt{7} \).Case 2: \( 2 \leq x<3 \)In this interval, \( |x - 3| = 3 - x \) and \( |x - 2| = x - 2 \). The equation becomes:\[x (3 - x) + 3 (x - 2) + 1 = 0\]\[3x - x^2 + 3x - 6 + 1 = 0\]\[-x^2 + 6x - 5 = 0\]Solving this quadratic equation yields:\[x = \frac{-6 \pm \sqrt{6^2 - 4(-1)(-5)}}{2(-1)} = \frac{-6 \pm \sqrt{16}}{-2} = \frac{-6 \pm 4}{-2}\]The solutions are \( x = 1 \) (invalid as \( x \geq 2 \)) and \( x = 5 \) (invalid as \( x<3 \)). No valid solution exists in this case.Case 3: \( x \geq 3 \)Here, \( |x - 3| = x - 3 \) and \( |x - 2| = x - 2 \). The equation transforms to:\[x (x - 3) + 3 (x - 2) + 1 = 0\]\[x^2 - 3x + 3x - 6 + 1 = 0\]\[x^2 - 5 = 0\]\[x^2 = 5\]\[x = \pm \sqrt{5}\]Considering \( x \geq 3 \), the valid solution is \( x = \sqrt{5} \). Therefore, the equation has one valid real solution: \( x = -\sqrt{7} \).