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

We are given the equation: \[ x |x - 3| + 3 |x - 2| + 1 = 0 \] To solve for \( x \), we need to consider different cases based on the values of \( x \) because of the absolute value expressions. The points where the expressions inside the absolute values change their sign are \( x = 3 \) and \( x = 2 \). Thus, we divide the solution into three intervals: \( x<2 \), \( 2 \leq x<3 \), and \( x \geq 3 \). Case 1: \( x<2 \)
In this case, both \( |x - 3| = 3 - x \) and \( |x - 2| = 2 - x \). The equation becomes: \[ x (3 - x) + 3 (2 - x) + 1 = 0 \] Expanding: \[ x (3 - x) = 3x - x^2 \quad \text{and} \quad 3 (2 - x) = 6 - 3x \] Thus, the equation becomes: \[ 3x - x^2 + 6 - 3x + 1 = 0 \] Simplifying: \[ -x^2 + 7 = 0 \] \[ x^2 = 7 \] \[ x = \pm \sqrt{7} \] Since \( x<2 \), we take the solution \( x = -\sqrt{7} \). Case 2: \( 2 \leq x<3 \)
In this case, \( |x - 3| = 3 - x \) and \( |x - 2| = x - 2 \). The equation becomes: \[ x (3 - x) + 3 (x - 2) + 1 = 0 \] Expanding: \[ x (3 - x) = 3x - x^2 \quad \text{and} \quad 3 (x - 2) = 3x - 6 \] Thus, the equation becomes: \[ 3x - x^2 + 3x - 6 + 1 = 0 \] Simplifying: \[ - x^2 + 6x - 5 = 0 \] Solving the quadratic equation: \[ x = \frac{-6 \pm \sqrt{6^2 - 4(-1)(-5)}}{2(-1)} = \frac{-6 \pm \sqrt{36 - 20}}{-2} = \frac{-6 \pm \sqrt{16}}{-2} \] \[ x = \frac{-6 \pm 4}{-2} \] This gives two solutions: \[ x = \frac{-6 + 4}{-2} = 1 \quad \text{(Not valid as \( x \geq 2 \))} \] \[ x = \frac{-6 - 4}{-2} = 5 \] Since \( 5 \) is not in the interval \( 2 \leq x<3 \), no valid solution occurs in this case. Case 3: \( x \geq 3 \)
In this case, both \( |x - 3| = x - 3 \) and \( |x - 2| = x - 2 \). The equation becomes: \[ x (x - 3) + 3 (x - 2) + 1 = 0 \] Expanding: \[ x (x - 3) = x^2 - 3x \quad \text{and} \quad 3 (x - 2) = 3x - 6 \] Thus, the equation becomes: \[ x^2 - 3x + 3x - 6 + 1 = 0 \] Simplifying: \[ x^2 - 5 = 0 \] \[ x^2 = 5 \] \[ x = \pm \sqrt{5} \] Since \( x \geq 3 \), we take the solution \( x = \sqrt{5} \). Thus, there is only one valid real solution: \( x = -\sqrt{7} \).