Key Concepts and Formulas

• Definition of Absolute Value: For any real number $a$, $|a| = a$ if $a \ge 0$, and $|a| = -a$ if $a < 0$.
• Solving Absolute Value Equations: Equations involving absolute values are typically solved by considering different cases based on the sign of the expression inside the absolute value. The points where the expression equals zero are called critical points, and they divide the number line into intervals.
• 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, $\Delta = b^2 - 4ac$, determines the nature of the roots: if $\Delta < 0$, there are no real roots.

Step-by-Step Solution

Step 1: Identify critical points and intervals. The given equation is $(|x| - 3)|x + 4| = 6$. We have two absolute value expressions: $|x|$ and $|x+4|$. The critical points are where the expressions inside the absolute values become zero. For $|x|$, the critical point is $x=0$. For $|x+4|$, the critical point is $x+4=0$, which means $x=-4$. These critical points, $x=-4$ and $x=0$, divide the real number line into three intervals: $(-\infty, -4)$, $[-4, 0)$, and $[0, \infty)$. We will analyze the equation in each of these intervals.

Step 2: Analyze Case 1: $x < -4$. In this interval, $x$ is negative, so $|x| = -x$. Also, $x+4$ is negative, so $|x+4| = -(x+4)$. Substitute these into the original equation: $(-x - 3) (-(x+4)) = 6$ $(x+3)(x+4) = 6$ Expand the left side: $x^2 + 4x + 3x + 12 = 6$ $x^2 + 7x + 12 = 6$ Rearrange into a standard quadratic equation: $x^2 + 7x + 6 = 0$ Factor the quadratic equation: $(x+1)(x+6) = 0$ This gives potential solutions $x=-1$ and $x=-6$. We must check these against the condition $x < -4$. $x=-1$ does not satisfy $x < -4$. $x=-6$ satisfies $x < -4$. So, $x=-6$ is a valid solution from this case.

Step 3: Analyze Case 2: $-4 \le x < 0$. In this interval, $x$ is negative, so $|x| = -x$. However, $x+4$ is non-negative (since $x \ge -4$), so $|x+4| = x+4$. Substitute these into the original equation: $(-x - 3) (x+4) = 6$ Expand the left side: $-x^2 - 4x - 3x - 12 = 6$ $-x^2 - 7x - 12 = 6$ Rearrange into a standard quadratic equation: $x^2 + 7x + 18 = 0$ To find the solutions, we calculate the discriminant $\Delta = b^2 - 4ac$: $\Delta = (7)^2 - 4(1)(18) = 49 - 72 = -23$ Since $\Delta < 0$, there are no real solutions in this interval.

Step 4: Analyze Case 3: $x \ge 0$. In this interval, $x$ is non-negative, so $|x| = x$. Also, $x+4$ is positive, so $|x+4| = x+4$. Substitute these into the original equation: $(x - 3) (x+4) = 6$ Expand the left side: $x^2 + 4x - 3x - 12 = 6$ $x^2 + x - 12 = 6$ Rearrange into a standard quadratic equation: $x^2 + x - 18 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1, b=1, c=-18$: $x = \frac{-1 \pm \sqrt{(1)^2 - 4(1)(-18)}}{2(1)}$ $x = \frac{-1 \pm \sqrt{1 + 72}}{2}$ $x = \frac{-1 \pm \sqrt{73}}{2}$ This gives two potential solutions: $x_1 = \frac{-1 + \sqrt{73}}{2}$ and $x_2 = \frac{-1 - \sqrt{73}}{2}$. We must check these against the condition $x \ge 0$. For $x_1 = \frac{-1 + \sqrt{73}}{2}$: Since $\sqrt{73}$ is between $\sqrt{64}=8$ and $\sqrt{81}=9$, $\sqrt{73} \approx 8.5$. Thus, $\frac{-1 + \sqrt{73}}{2} > 0$. This solution satisfies $x \ge 0$. For $x_2 = \frac{-1 - \sqrt{73}}{2}$: This value is clearly negative, so it does not satisfy $x \ge 0$. So, $x = \frac{-1 + \sqrt{73}}{2}$ is a valid solution from this case.

Step 5: Consolidate the solutions. The valid solutions found from the three cases are: From Case 1: $x = -6$ From Case 2: No real solutions From Case 3: $x = \frac{-1 + \sqrt{73}}{2}$ The set of solutions is $\left\{ -6, \frac{-1 + \sqrt{73}}{2} \right\}$. The number of elements in this set is 2.

Common Mistakes & Tips

• Forgetting to check solutions against the case condition: Always verify if a potential solution falls within the interval defined for that case. Extraneous solutions must be discarded.
• Algebraic errors with signs: Pay close attention to the signs when expanding products and rearranging terms, especially when dealing with negative numbers in absolute values.
• Incorrectly determining the sign of expressions within absolute values: Double-check the sign of $|x|$ and $|x+4|$ in each interval.

Summary

The problem of solving the equation $(|x| - 3)|x + 4| = 6$ was approached by breaking it down into three cases based on the critical points $x=-4$ and $x=0$. In each case, the absolute value expressions were simplified, leading to standard algebraic equations. After solving these equations, the potential solutions were rigorously checked against the conditions of their respective cases. This process yielded two valid real solutions: $x = -6$ and $x = \frac{-1 + \sqrt{73}}{2}$. Therefore, the set of elements satisfying the equation contains two numbers.

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