Key Concepts and Formulas

• 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 $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Case Analysis: Divide the domain into intervals based on where the expressions inside the absolute values change signs.

Step-by-Step Solution

Step 1: Identify Critical Points

We need to find the values of $x$ where the expressions inside the absolute value signs change their signs.

• $|x|$ changes sign at $x = 0$.
• $|x+2|$ changes sign at $x+2 = 0$, which means $x = -2$.

These critical points, $x = -2$ and $x = 0$, divide the real number line into three intervals: $x < -2$, $-2 \le x < 0$, and $x \ge 0$.

Step 2: Case 1: $x < -2$

Why? In this interval, both $x$ and $x+2$ are negative.

Since $x < -2$, we have $x < 0$ and $x+2 < 0$. Therefore, $|x| = -x$ and $|x+2| = -(x+2)$. Substituting into the original equation: $x|x| - 5|x+2| + 6 = 0$ $x(-x) - 5(-(x+2)) + 6 = 0$ $-x^2 + 5(x+2) + 6 = 0$ $-x^2 + 5x + 10 + 6 = 0$ $-x^2 + 5x + 16 = 0$ Multiplying by $-1$, we get: $x^2 - 5x - 16 = 0$ Using the quadratic formula: $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-16)}}{2(1)}$ $x = \frac{5 \pm \sqrt{25 + 64}}{2}$ $x = \frac{5 \pm \sqrt{89}}{2}$

Now we need to check if these roots satisfy the condition $x < -2$. $x_1 = \frac{5 + \sqrt{89}}{2} \approx \frac{5 + 9.43}{2} \approx 7.21$ Since $7.21 > -2$, this root is not valid. $x_2 = \frac{5 - \sqrt{89}}{2} \approx \frac{5 - 9.43}{2} \approx -2.21$ Since $-2.21 < -2$, this root is valid.

Step 3: Case 2: $-2 \le x < 0$

Why? In this interval, $x$ is negative or zero, but $x+2$ is non-negative.

Since $-2 \le x < 0$, we have $|x| = -x$ and $|x+2| = x+2$. Substituting into the original equation: $x|x| - 5|x+2| + 6 = 0$ $x(-x) - 5(x+2) + 6 = 0$ $-x^2 - 5x - 10 + 6 = 0$ $-x^2 - 5x - 4 = 0$ Multiplying by $-1$, we get: $x^2 + 5x + 4 = 0$ Factoring the quadratic: $(x+1)(x+4) = 0$ So, $x = -1$ or $x = -4$.

Now we need to check if these roots satisfy the condition $-2 \le x < 0$. $x = -1$ satisfies the condition since $-2 \le -1 < 0$. $x = -4$ does not satisfy the condition since $-4 < -2$.

Step 4: Case 3: $x \ge 0$

Why? In this interval, both $x$ and $x+2$ are non-negative.

Since $x \ge 0$, we have $|x| = x$ and $|x+2| = x+2$. Substituting into the original equation: $x|x| - 5|x+2| + 6 = 0$ $x(x) - 5(x+2) + 6 = 0$ $x^2 - 5x - 10 + 6 = 0$ $x^2 - 5x - 4 = 0$ Using the quadratic formula: $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-4)}}{2(1)}$ $x = \frac{5 \pm \sqrt{25 + 16}}{2}$ $x = \frac{5 \pm \sqrt{41}}{2}$

Now we need to check if these roots satisfy the condition $x \ge 0$. $x_3 = \frac{5 + \sqrt{41}}{2} \approx \frac{5 + 6.4}{2} \approx 5.7$ Since $5.7 \ge 0$, this root is valid. $x_4 = \frac{5 - \sqrt{41}}{2} \approx \frac{5 - 6.4}{2} \approx -0.7$ Since $-0.7 < 0$, this root is not valid.

Step 5: Count the Valid Roots

We found three valid real roots:

• $x = \frac{5 - \sqrt{89}}{2}$ from Case 1
• $x = -1$ from Case 2
• $x = \frac{5 + \sqrt{41}}{2}$ from Case 3

Therefore, there are 3 real roots in total.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when substituting the absolute value expressions in each case.
• Interval Check: Always verify that the roots obtained in each case satisfy the condition of that particular interval.
• Quadratic Formula: Double check the quadratic formula application.

Summary

The problem was solved by considering three cases, each corresponding to a different interval determined by the points where the expressions inside the absolute values change signs. In each case, the absolute value expressions were replaced with their appropriate linear forms, resulting in quadratic equations. The solutions to these quadratic equations were then checked against the interval conditions to ensure their validity. The total number of real roots is 3.

Final Answer

The final answer is \boxed{3}, which corresponds to option (B).