The number of real roots of the equation $x\lvert x - 2\rvert + 3\lvert x - 3\rvert + 1 = 0$ is:

Step 1: Define cases based on absolute values

To solve an equation involving absolute values, we need to consider different cases based on when the expressions inside the absolute values change sign. The critical points for this equation are $x=2$ (from $|x-2|$) and $x=3$ (from $|x-3|$). These points divide the number line into three intervals: $x < 2$, $2 \le x < 3$, and $x \ge 3$.

Step 2: Case 1: $x < 2$

In this interval, both $(x-2)$ and $(x-3)$ are negative. Therefore, $|x-2| = -(x-2)$ and $|x-3| = -(x-3)$. Substituting these into the original equation gives us a quadratic equation. We then solve this quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step 3: Solve for roots in Case 1

Applying the quadratic formula, we find two potential roots: $x = \frac{-1 + \sqrt{41}}{2}$ and $x = \frac{-1 - \sqrt{41}}{2}$. We need to check if these roots fall within our current interval $x < 2$. Since $\sqrt{41}$ is approximately 6.4, $\frac{-1 + 6.4}{2} \approx 2.7$ (not less than 2) and $\frac{-1 - 6.4}{2} \approx -3.7$ (which is less than 2). Thus, only $x = \frac{-1 - \sqrt{41}}{2}$ is a valid root in this case.

Step 4: Case 2: $2 \le x < 3$

In this interval, $(x-2)$ is non-negative, so $|x-2| = x-2$. However, $(x-3)$ is still negative, so $|x-3| = -(x-3)$. Substituting these into the equation yields another quadratic equation. We will again use the quadratic formula to find its roots.

Step 5: Solve for roots in Case 2

Applying the quadratic formula, we find that the discriminant $b^2 - 4ac = 25 - 40 = -15$, which is negative. A negative discriminant means there are no real roots in this interval. Therefore, there are no solutions for $x$ when $2 \le x < 3$.

Step 6: Case 3: $x \ge 3$

In this interval, both $(x-2)$ and $(x-3)$ are non-negative. So, $|x-2| = x-2$ and $|x-3| = x-3$. Substituting these into the original equation gives us the quadratic equation $x^2 + x - 8 = 0$. We will solve this using the quadratic formula.

Step 7: Solve for roots in Case 3

Applying the quadratic formula, we get two potential roots: $x = \frac{-1 + \sqrt{33}}{2}$ and $x = \frac{-1 - \sqrt{33}}{2}$. We check these against our condition $x \ge 3$. Since $\sqrt{33}$ is approximately 5.7, $\frac{-1 + 5.7}{2} \approx 2.35$ (not greater than or equal to 3) and $\frac{-1 - 5.7}{2} \approx -3.35$ (not greater than or equal to 3). Therefore, neither of these roots is valid in this interval.

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