Question

The number of real solution(s) of the equation \( x^2 + 3x + 2 = \min \left( |x - 3|, |x + 2| \right) \) is:

• A. \( 1 \)
• B. \( 3 \)
• C. \( 0 \)
• D. \( 2 \)

Hint:

When solving equations involving the minimum of absolute values, break the problem into cases based on the behavior of the absolute values and solve each case separately.

Answer 1

The Correct Option is A

We are tasked with solving the equation \( x^2 + 3x + 2 = \min \left( |x - 3|, |x + 2| \right) \). First, we analyze the behavior of the minimum function, which requires us to consider the cases for \( |x - 3| \) and \( |x + 2| \).

After checking these cases, we find that the equation has exactly one real solution.

Final Answer: \( 1 \).

Answer 2

Step 1: Analyze the equation \( x^2 + 3x + 2 = \min \left( |x - 3|, |x + 2| \right) \).
There will be two cases for the minimum function:
Case-1: \( x^2 + 3x + 2 = |x - 3| \) when \( |x - 3| < |x + 2| \)
Case-2: \( x^2 + 3x + 2 = |x + 2| \) when \( |x + 2| < |x - 3| \)

Step 2: Solve Case-1.
The inequality \( |x - 3| < |x + 2| \) holds when \( x < \frac{1}{2} \). For \( x < 3 \), \( |x - 3| = 3 - x \). Thus:
\[ x^2 + 3x + 2 = 3 - x \implies x^2 + 4x - 1 = 0 \] Solving this quadratic equation gives roots:
\[ x = -2 \pm \sqrt{5} \] Approximations are \( 0.236 \) and \( -4.236 \).
Checking the inequality \( |x - 3| < |x + 2| \) at these roots:
- At \( 0.236 \), \( |x - 3| = 2.764 \) and \( |x + 2| = 2.236 \), so \( |x - 3| > |x + 2| \), no solution.
- At \( -4.236 \), \( |x - 3| = 7.236 \) and \( |x + 2| = 2.236 \), no solution.

Step 3: Solve Case-2.
The inequality \( |x + 2| < |x - 3| \) holds when \( x > \frac{1}{2} \). For \( x \geq -2 \), \( |x + 2| = x + 2 \). Thus:
\[ x^2 + 3x + 2 = x + 2 \implies x^2 + 2x = 0 \implies x(x + 2) = 0 \] Roots \( x = 0 \) and \( x = -2 \) are not in the domain \( x > \frac{1}{2} \).
For \( x < -2 \), \( |x + 2| = -(x + 2) \) and:
\[ x^2 + 3x + 2 = -(x + 2) \implies x^2 + 4x + 4 = 0 \implies (x+2)^2 = 0 \implies x = -2 \] Again, \( x = -2 \) is not in the domain \( x > \frac{1}{2} \). No valid solutions.

Step 4: Check \( x = -2 \) explicitly.
At \( x = -2 \):
\[ x^2 + 3x + 2 = 0, \quad |x - 3| = 5, \quad |x + 2| = 0 \] Hence, \[ \min(|x - 3|, |x + 2|) = 0 \] So the equation holds true, giving one real solution.

Step 5: Final answer.
There is exactly one real solution.
\[ \boxed{1} \]