Key Concepts and Formulas

• Absolute Value Definition: $|a| = a$ if $a \ge 0$ and $|a| = -a$ if $a < 0$.
• Solving Equations with Absolute Values: Break down the problem into cases based on the intervals defined by the critical points of the expressions within the absolute value.
• Quadratic Formula: For a quadratic equation $ax^2 + bx + c = 0$, the solutions are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

Step 1: Identify Critical Points

The critical points are the values of $x$ that make the expressions inside the absolute values equal to zero.

• $x+5 = 0 \implies x = -5$
• $x+7 = 0 \implies x = -7$

These critical points divide the real number line into three intervals: $x < -7$, $-7 \le x < -5$, and $x \ge -5$.

Step 2: Case 1: $x \ge -5$

In this interval, $x+5 \ge 0$ and $x+7 \ge 0$. Therefore, $|x+5| = x+5$ and $|x+7| = x+7$. Substitute these into the original equation: $x(x+5) + 2(x+7) - 2 = 0$ Expand and simplify: $x^2 + 5x + 2x + 14 - 2 = 0$ $x^2 + 7x + 12 = 0$ Factor the quadratic: $(x+3)(x+4) = 0$ The potential solutions are $x = -3$ and $x = -4$.

Step 3: Verify Solutions for Case 1

Check if the potential solutions satisfy the condition $x \ge -5$:

• $x = -3$: $-3 \ge -5$ (True)
• $x = -4$: $-4 \ge -5$ (True) Both $x = -3$ and $x = -4$ are valid solutions in this case.

Step 4: Case 2: $-7 \le x < -5$

In this interval, $x+5 < 0$ and $x+7 \ge 0$. Therefore, $|x+5| = -(x+5)$ and $|x+7| = x+7$. Substitute these into the original equation: $x(-(x+5)) + 2(x+7) - 2 = 0$ Expand and simplify: $-x^2 - 5x + 2x + 14 - 2 = 0$ $-x^2 - 3x + 12 = 0$ Multiply by $-1$: $x^2 + 3x - 12 = 0$ Use the quadratic formula: $x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-12)}}{2(1)}$ $x = \frac{-3 \pm \sqrt{9 + 48}}{2}$ $x = \frac{-3 \pm \sqrt{57}}{2}$ The potential solutions are $x = \frac{-3 + \sqrt{57}}{2}$ and $x = \frac{-3 - \sqrt{57}}{2}$.

Step 5: Verify Solutions for Case 2

Check if the potential solutions satisfy the condition $-7 \le x < -5$:

• $x = \frac{-3 + \sqrt{57}}{2} \approx \frac{-3 + 7.55}{2} \approx 2.275$. This is not in the interval $[-7, -5)$.
• $x = \frac{-3 - \sqrt{57}}{2} \approx \frac{-3 - 7.55}{2} \approx -5.275$. This is in the interval $[-7, -5)$. Only $x = \frac{-3 - \sqrt{57}}{2}$ is a valid solution in this case.

Step 6: Case 3: $x < -7$

In this interval, $x+5 < 0$ and $x+7 < 0$. Therefore, $|x+5| = -(x+5)$ and $|x+7| = -(x+7)$. Substitute these into the original equation: $x(-(x+5)) + 2(-(x+7)) - 2 = 0$ Expand and simplify: $-x^2 - 5x - 2x - 14 - 2 = 0$ $-x^2 - 7x - 16 = 0$ Multiply by $-1$: $x^2 + 7x + 16 = 0$ Calculate the discriminant: $D = 7^2 - 4(1)(16) = 49 - 64 = -15$ Since $D < 0$, there are no real solutions in this case.

Step 7: Count the Real Solutions

• Case 1: 2 real solutions ($x=-3, x=-4$)
• Case 2: 1 real solution ($x = \frac{-3 - \sqrt{57}}{2}$)
• Case 3: 0 real solutions

Total number of real solutions: $2 + 1 + 0 = 3$.

Common Mistakes & Tips

• Sign Errors: Double-check the signs when substituting and simplifying expressions with absolute values.
• Interval Verification: Always verify that the solutions obtained in each case satisfy the interval's condition. Failing to do so can lead to extraneous solutions.
• Approximation: When dealing with square roots, approximating their values can help determine if they fall within the correct interval.

Summary

By analyzing the equation $x|x+5|+2|x+7|-2=0$ in three distinct intervals, we found two real solutions in the first interval, one real solution in the second interval, and no real solutions in the third interval. Therefore, the equation has a total of 3 real solutions.

Final Answer The final answer is \boxed{3}.