Question

The number of solution(s) of the equation \[ x |x + 4| + 3 |x + 2| = 0 \] is/are equal to:

Hint:

When solving absolute value equations, break the problem into cases based on the sign of the expression inside the absolute value.

Answer 1

Correct Answer: 3

To solve the equation \( x |x + 4| + 3 |x + 2| = 0 \), we need to consider the absolute values, which lead to different cases based on the values of \( x \).

Case 1: \( x \geq -2 \)
Here, both \( |x + 4| = x + 4 \) and \( |x + 2| = x + 2 \). Substituting these into the equation gives:
\( x(x + 4) + 3(x + 2) = 0 \)
Expanding, we have:
\( x^2 + 4x + 3x + 6 = 0 \)
\( x^2 + 7x + 6 = 0 \)
To factor this quadratic equation, we find two numbers that multiply to 6 and add to 7: these are 6 and 1. So, the equation factors as:
\( (x + 6)(x + 1) = 0 \)
Thus, \( x = -6 \) or \( x = -1 \). However, \( x \geq -2 \) limits our solution to \( x = -1 \).

Case 2: \( -4 \leq x < -2 \)
In this range, \( |x + 4| = x + 4 \) and \( |x + 2| = -(x + 2) \). Substituting these gives:
\( x(x + 4) + 3(-x - 2) = 0 \)
Expanding, we have:
\( x^2 + 4x - 3x - 6 = 0 \)
\( x^2 + x - 6 = 0 \)
To factor \( x^2 + x - 6 \), we need two numbers that multiply to -6 and add to 1: these are 3 and -2. Hence:
\( (x + 3)(x - 2) = 0 \)
This gives \( x = -3 \) or \( x = 2 \). Only \( x = -3 \) falls within the range.

Case 3: \( x < -4 \)
Here, both \( |x + 4| = -(x + 4) \) and \( |x + 2| = -(x + 2) \). Substituting, we have:
\( x(-x - 4) + 3(-x - 2) = 0 \)
Expanding, we get:
\( -x^2 - 4x - 3x - 6 = 0 \)
\( -x^2 - 7x - 6 = 0 \) or equivalently, \( x^2 + 7x + 6 = 0 \)
The factors \( (x + 6)(x + 1) = 0 \) yield \( x = -6 \) or \( x = -1 \). Only \( x = -6 \) falls in this range.

Solution Summary:
The solutions are \( x = -6, -3, -1 \), totaling 3 solutions.
This agrees with the expected range (3, 3).