Key Concepts and Formulas

• Zero Product Property: If $A \cdot B = 0$, then $A = 0$ or $B = 0$ (or both).
• 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 $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

Step 1: Apply the Zero Product Property

We are given the equation $x\left(x^2+3|x|+5|x-1|+6|x-2|\right)=0.$ Using the Zero Product Property, we can split this into two separate equations: $x = 0$ $x^2+3|x|+5|x-1|+6|x-2|=0$ We will analyze each equation separately.

Step 2: Analyze the first equation, $x=0$

The first equation is simply $x=0$. We need to check if this solution is valid. Substituting $x=0$ into the original equation: $0\left(0^2+3|0|+5|0-1|+6|0-2|\right) = 0$ Since this is true, $x=0$ is a real solution.

Step 3: Analyze the second equation, $x^2+3|x|+5|x-1|+6|x-2|=0$

To solve $x^2+3|x|+5|x-1|+6|x-2|=0$, we need to consider different cases based on the values of $x$ that make the expressions inside the absolute values equal to zero. These critical points are $x=0$, $x=1$, and $x=2$. These points divide the real number line into four intervals: $x<0$, $0 \le x < 1$, $1 \le x < 2$, and $x \ge 2$. We will analyze each interval separately.

Step 4: Case A: $x < 0$

In this interval, $|x| = -x$, $|x-1| = 1-x$, and $|x-2| = 2-x$. Substituting these into the equation gives: $x^2 + 3(-x) + 5(1-x) + 6(2-x) = 0$ $x^2 - 3x + 5 - 5x + 12 - 6x = 0$ $x^2 - 14x + 17 = 0$ We use the quadratic formula to solve for $x$: $x = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(1)(17)}}{2(1)} = \frac{14 \pm \sqrt{196 - 68}}{2} = \frac{14 \pm \sqrt{128}}{2} = \frac{14 \pm 8\sqrt{2}}{2} = 7 \pm 4\sqrt{2}$ So $x = 7 + 4\sqrt{2} \approx 7 + 4(1.414) \approx 12.656$ and $x = 7 - 4\sqrt{2} \approx 7 - 4(1.414) \approx 1.344$. Since we are considering the interval $x < 0$, and both of these values are greater than 0, neither is a valid solution in this interval.

Step 5: Case B: $0 \le x < 1$

In this interval, $|x| = x$, $|x-1| = 1-x$, and $|x-2| = 2-x$. Substituting these into the equation gives: $x^2 + 3(x) + 5(1-x) + 6(2-x) = 0$ $x^2 + 3x + 5 - 5x + 12 - 6x = 0$ $x^2 - 8x + 17 = 0$ The discriminant is $D = b^2 - 4ac = (-8)^2 - 4(1)(17) = 64 - 68 = -4$. Since the discriminant is negative, there are no real solutions in this interval.

Step 6: Case C: $1 \le x < 2$

In this interval, $|x| = x$, $|x-1| = x-1$, and $|x-2| = 2-x$. Substituting these into the equation gives: $x^2 + 3(x) + 5(x-1) + 6(2-x) = 0$ $x^2 + 3x + 5x - 5 + 12 - 6x = 0$ $x^2 + 2x + 7 = 0$ The discriminant is $D = b^2 - 4ac = (2)^2 - 4(1)(7) = 4 - 28 = -24$. Since the discriminant is negative, there are no real solutions in this interval.

Step 7: Case D: $x \ge 2$

In this interval, $|x| = x$, $|x-1| = x-1$, and $|x-2| = x-2$. Substituting these into the equation gives: $x^2 + 3(x) + 5(x-1) + 6(x-2) = 0$ $x^2 + 3x + 5x - 5 + 6x - 12 = 0$ $x^2 + 14x - 17 = 0$ We use the quadratic formula to solve for $x$: $x = \frac{-14 \pm \sqrt{14^2 - 4(1)(-17)}}{2(1)} = \frac{-14 \pm \sqrt{196 + 68}}{2} = \frac{-14 \pm \sqrt{264}}{2} = \frac{-14 \pm 2\sqrt{66}}{2} = -7 \pm \sqrt{66}$ So $x = -7 + \sqrt{66} \approx -7 + 8.12 \approx 1.12$ and $x = -7 - \sqrt{66} \approx -7 - 8.12 \approx -15.12$. Since we are considering the interval $x \ge 2$, and both of these values are less than 2, neither is a valid solution in this interval.

Step 8: Consolidate Solutions

The only real solution we found is $x=0$.

Common Mistakes & Tips

• Remember to check if the solutions obtained in each case satisfy the initial condition of that case.
• Pay close attention to the signs when dealing with absolute values.
• The discriminant can quickly tell you if a quadratic has real roots.

Summary

We used the Zero Product Property to split the original equation into two simpler equations. We then analyzed the second equation by considering different cases based on the critical points of the absolute value expressions. After checking each case, we found that the only real solution to the original equation is $x=0$. Therefore, the number of real solutions is 1.

Final Answer The final answer is \boxed{1}.