Key Concepts and Formulas

• Solving Equations with Minimum Functions: To solve $f(x) = \min\{g(x), h(x)\}$, analyze where $g(x) = h(x)$ to determine the intervals where $g(x)$ or $h(x)$ defines the minimum.
• Absolute Value Definition: $|x| = x$ if $x \ge 0$, and $|x| = -x$ if $x < 0$.
• Quadratic Formula: For a quadratic equation $ax^2 + bx + c = 0$, the solutions are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. The discriminant is $D = b^2 - 4ac$. If $D < 0$, the quadratic has no real roots.

Step-by-Step Solution

Step 1: Analyze the given equation and identify critical points

The given equation is $x^2 + 3x + 2 = \min\{|x-3|, |x+2|\}$. We need to find the real solutions to this equation. The critical points for the absolute value functions are $x = 3$ and $x = -2$. Also, we need to find where $|x-3| = |x+2|$.

Step 2: Find the point where the absolute values are equal

We solve $|x-3| = |x+2|$. This is equivalent to $x-3 = x+2$ or $x-3 = -(x+2)$. The first equation, $x-3 = x+2$, gives $-3=2$, which is impossible, so there is no solution. The second equation, $x-3 = -x-2$, gives $2x = 1$, so $x = \frac{1}{2}$.

Step 3: Define the piecewise minimum function

We have critical points at $x=-2, x=1/2,$ and $x=3$. These divide the real line into intervals. However, we only need to consider where $|x-3|$ is less than or equal to $|x+2|$ or vice versa. We know the switch happens at $x = 1/2$.

• If $x < \frac{1}{2}$, then $|x+2| < |x-3|$, so $\min\{|x-3|, |x+2|\} = |x+2|$.
• If $x \ge \frac{1}{2}$, then $|x-3| \le |x+2|$, so $\min\{|x-3|, |x+2|\} = |x-3|$.

Therefore, the piecewise definition of $\min\{|x-3|, |x+2|\}$ is:

$$\min\{|x-3|, |x+2|\} = \begin{cases} |x+2| = -x-2, & x < -2 \\ |x+2| = x+2, & -2 \le x < \frac{1}{2} \\ |x-3| = 3-x, & \frac{1}{2} \le x < 3 \\ |x-3| = x-3, & x \ge 3 \end{cases}$$

Step 4: Solve the equation in each interval

• Case 1: $x < -2$ $x^2 + 3x + 2 = -x - 2$ $x^2 + 4x + 4 = 0$ $(x+2)^2 = 0$ $x = -2$. Since $x < -2$, this solution is not valid.

• Case 2: $-2 \le x < \frac{1}{2}$ $x^2 + 3x + 2 = x + 2$ $x^2 + 2x = 0$ $x(x+2) = 0$ $x = 0$ or $x = -2$. Both solutions are in the interval $[-2, \frac{1}{2})$. So, $x=0$ and $x=-2$ are valid solutions.

• Case 3: $\frac{1}{2} \le x < 3$ $x^2 + 3x + 2 = 3 - x$ $x^2 + 4x - 1 = 0$ $x = \frac{-4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{-4 \pm \sqrt{20}}{2} = \frac{-4 \pm 2\sqrt{5}}{2} = -2 \pm \sqrt{5}$ $x = -2 + \sqrt{5} \approx -2 + 2.236 = 0.236$ or $x = -2 - \sqrt{5} \approx -2 - 2.236 = -4.236$ Since $\frac{1}{2} \le x < 3$, we check if the solutions are in this interval. $x=-2+\sqrt{5} \approx 0.236$ is not in the interval $[\frac{1}{2}, 3)$. $x = -2 - \sqrt{5}$ is not in the interval. So there are no valid solutions in this interval.

• Case 4: $x \ge 3$ $x^2 + 3x + 2 = x - 3$ $x^2 + 2x + 5 = 0$ $D = 2^2 - 4(1)(5) = 4 - 20 = -16 < 0$. Therefore, there are no real solutions in this interval.

Step 5: Count the valid solutions

The only valid solutions are $x = 0$ and $x = -2$. Therefore, there are 2 real solutions.

Common Mistakes & Tips

• Forgetting to check intervals: Always verify that your solutions lie within the intervals you defined.
• Incorrectly solving absolute values: Remember to consider both positive and negative cases when dealing with absolute values.
• Assuming the minimum switch point is obvious: The minimum switch point is where the two absolute values are equal, not necessarily at one of the critical points of the absolute value functions.

Summary

By analyzing the piecewise function defined by the minimum of the absolute value expressions and solving the resulting quadratic equations in each interval, we found two real solutions: $x=0$ and $x=-2$. Thus, the equation has 2 real solutions.

Final Answer

The final answer is \boxed{2}, which corresponds to option (A).