Key Concepts and Formulas

• Absolute Value Definition: $|x| = x$ if $x \geq 0$, and $|x| = -x$ if $x < 0$.
• Critical Points: The values of $x$ where the expression inside an absolute value equals zero. These values divide the number line into intervals where the absolute value can be expressed without the absolute value sign.
• Quadratic Formula: For a quadratic equation $ax^2 + bx + c = 0$, the solutions are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Discriminant: For a quadratic equation $ax^2 + bx + c = 0$, the discriminant is $D = b^2 - 4ac$. If $D < 0$, the equation has no real roots.

Step-by-Step Solution

Step 1: Identify the Critical Points

The given equation is $x |x - 2| + 3|x - 3| + 1 = 0$. The expressions inside the absolute values are $x - 2$ and $x - 3$. Setting these equal to zero, we find the critical points:

• $x - 2 = 0 \Rightarrow x = 2$
• $x - 3 = 0 \Rightarrow x = 3$

These critical points divide the real number line into three intervals: $x < 2$, $2 \leq x < 3$, and $x \geq 3$.

Step 2: Analyze Case (I): $x < 2$

In this interval, $x - 2 < 0$ and $x - 3 < 0$. Therefore, $|x - 2| = -(x - 2) = 2 - x$ and $|x - 3| = -(x - 3) = 3 - x$. Substituting these into the original equation:

$x(2 - x) + 3(3 - x) + 1 = 0$ $2x - x^2 + 9 - 3x + 1 = 0$ $-x^2 - x + 10 = 0$ $x^2 + x - 10 = 0$

Using the quadratic formula with $a = 1$, $b = 1$, and $c = -10$:

$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-10)}}{2(1)}$ $x = \frac{-1 \pm \sqrt{41}}{2}$

We have two potential roots: $x_1 = \frac{-1 + \sqrt{41}}{2}$ and $x_2 = \frac{-1 - \sqrt{41}}{2}$. Since $\sqrt{41} \approx 6.4$,

$x_1 \approx \frac{-1 + 6.4}{2} = \frac{5.4}{2} = 2.7$. Since $2.7 \nless 2$, we reject this root. $x_2 \approx \frac{-1 - 6.4}{2} = \frac{-7.4}{2} = -3.7$. Since $-3.7 < 2$, we accept this root.

Thus, in this interval, we have one real root: $x = \frac{-1 - \sqrt{41}}{2}$.

Step 3: Analyze Case (II): $2 \leq x < 3$

In this interval, $x - 2 \geq 0$ and $x - 3 < 0$. Therefore, $|x - 2| = x - 2$ and $|x - 3| = -(x - 3) = 3 - x$. Substituting these into the original equation:

$x(x - 2) + 3(3 - x) + 1 = 0$ $x^2 - 2x + 9 - 3x + 1 = 0$ $x^2 - 5x + 10 = 0$

The discriminant is $D = (-5)^2 - 4(1)(10) = 25 - 40 = -15$. Since $D < 0$, this quadratic equation has no real roots.

Step 4: Analyze Case (III): $x \geq 3$

In this interval, $x - 2 > 0$ and $x - 3 \geq 0$. Therefore, $|x - 2| = x - 2$ and $|x - 3| = x - 3$. Substituting these into the original equation:

$x(x - 2) + 3(x - 3) + 1 = 0$ $x^2 - 2x + 3x - 9 + 1 = 0$ $x^2 + x - 8 = 0$

Using the quadratic formula with $a = 1$, $b = 1$, and $c = -8$:

$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-8)}}{2(1)}$ $x = \frac{-1 \pm \sqrt{33}}{2}$

We have two potential roots: $x_1 = \frac{-1 + \sqrt{33}}{2}$ and $x_2 = \frac{-1 - \sqrt{33}}{2}$. Since $\sqrt{33} \approx 5.74$,

$x_1 \approx \frac{-1 + 5.74}{2} = \frac{4.74}{2} = 2.37$. Since $2.37 \nless 3$, we reject this root. $x_2 \approx \frac{-1 - 5.74}{2} = \frac{-6.74}{2} = -3.37$. Since $-3.37 \nless 3$, we reject this root.

Thus, in this interval, we have no real roots.

Step 5: Count the Real Roots

Combining the results from all three cases:

• Case (I) $x < 2$: 1 real root
• Case (II) $2 \leq x < 3$: 0 real roots
• Case (III) $x \geq 3$: 0 real roots

Therefore, the total number of real roots for the equation $x |x - 2| + 3|x - 3| + 1 = 0$ is $1$.

Common Mistakes & Tips

• Remember to check if the calculated roots lie within the assumed interval. Reject any roots that do not satisfy the interval condition.
• Pay close attention to the signs when removing the absolute value signs. This is the most common source of errors.
• If the discriminant of a quadratic equation is negative, there are no real roots.

Summary

Solving the given equation involving absolute values requires considering different cases based on the critical points of the absolute value expressions. After correctly defining the absolute values in each interval, we solve the resulting quadratic equations. It is crucial to check if the roots obtained fall within the assumed interval. In this case, we find only one real root.

Final Answer

The final answer is $\boxed{1}$, which corresponds to option (D).