Solving the Absolute Value Equation

We need to find the number of real roots for the equation $x |x-2|+3|x-3|+1=0$. We will solve this by considering different cases based on the values of $x$ where the expressions inside the absolute value signs change sign.

The critical points are $x=2$ and $x=3$. These points divide the number line into three intervals: $x < 2$, $2 \le x < 3$, and $x \ge 3$. We analyze the equation in each interval.

Case Analysis

Case 1: $x < 2$

• In this interval, $|x-2| = -(x-2) = 2-x$ and $|x-3| = -(x-3) = 3-x$.
• The equation becomes: $x(2-x) + 3(3-x) + 1 = 0$.
• Simplifying: $2x - x^2 + 9 - 3x + 1 = 0$.
• This gives the quadratic equation: $-x^2 - x + 10 = 0$, or $x^2 + x - 10 = 0$.
• Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-10)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 40}}{2} = \frac{-1 \pm \sqrt{41}}{2}$.
• The two potential roots are $x_1 = \frac{-1 + \sqrt{41}}{2}$ and $x_2 = \frac{-1 - \sqrt{41}}{2}$.
• We check if these roots satisfy the condition $x < 2$.
  • $\sqrt{41}$ is between 6 and 7 (approx 6.4).
  • $x_1 \approx \frac{-1 + 6.4}{2} \approx 2.7$. This is not less than 2.
  • $x_2 \approx \frac{-1 - 6.4}{2} \approx -3.7$. This is less than 2.
• Therefore, $x = \frac{-1 - \sqrt{41}}{2}$ is a valid root.

Case 2: $2 \le x < 3$

• In this interval, $|x-2| = x-2$ and $|x-3| = -(x-3) = 3-x$.
• The equation becomes: $x(x-2) + 3(3-x) + 1 = 0$.
• Simplifying: $x^2 - 2x + 9 - 3x + 1 = 0$.
• This gives the quadratic equation: $x^2 - 5x + 10 = 0$.
• Calculate the discriminant $\Delta = b^2 - 4ac = (-5)^2 - 4(1)(10) = 25 - 40 = -15$.
• Since the discriminant $\Delta < 0$, there are no real roots in this interval.

Case 3: $x \ge 3$

• In this interval, $|x-2| = x-2$ and $|x-3| = x-3$.
• The equation becomes: $x(x-2) + 3(x-3) + 1 = 0$.
• Simplifying: $x^2 - 2x + 3x - 9 + 1 = 0$.
• This gives the quadratic equation: $x^2 + x - 8 = 0$.
• Using the quadratic formula: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-8)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 32}}{2} = \frac{-1 \pm \sqrt{33}}{2}$.
• The two potential roots are $x_3 = \frac{-1 + \sqrt{33}}{2}$ and $x_4 = \frac{-1 - \sqrt{33}}{2}$.
• We check if these roots satisfy the condition $x \ge 3$.
  • $\sqrt{33}$ is between 5 and 6 (approx 5.7).
  • $x_3 \approx \frac{-1 + 5.7}{2} \approx 2.35$. This is not greater than or equal to 3.
  • $x_4 \approx \frac{-1 - 5.7}{2} \approx -3.35$. This is not greater than or equal to 3.
• Therefore, there are no valid roots in this interval.

Conclusion

Combining the results from all three cases, we found only one real root: $x = \frac{-1 - \sqrt{41}}{2}$.

The number of real roots is 1.