1. Key Concepts and Formulas

• Discriminant of a Quadratic Equation: For a quadratic equation $ax^2 + bx + c = 0$, the discriminant is given by $D = b^2 - 4ac$.
• Condition for Distinct Real Roots: An equation $ax^2 + bx + c = 0$ has two distinct real roots if and only if its discriminant $D$ is strictly positive, i.e., $D > 0$. The phrase "one real root bigger than the other" explicitly implies two distinct real roots.
• Counting Principle for Sample Space: If there are $n_1$ choices for the first event, $n_2$ choices for the second, and $n_3$ choices for the third, then the total number of possible combinations is $n_1 \times n_2 \times n_3$.
• Probability Formula: The probability $p$ of an event is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes: $p = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.

2. Step-by-Step Solution

Step 1: Establish the condition for the quadratic equation to have distinct real roots. The problem states that the quadratic equation $ax^2 + bx + c = 0$ has "one real root bigger than the other". This is a precise way of saying that the equation must have two distinct real roots. For a quadratic equation, the nature of its roots is determined by its discriminant, $D = b^2 - 4ac$.

• If $D > 0$, there are two distinct real roots.
• If $D = 0$, there is exactly one real root (a repeated or double root).
• If $D < 0$, there are two distinct complex roots. Therefore, for the given condition, we require the discriminant to be strictly positive: $D = b^2 - 4ac > 0$ This inequality can be rewritten as $b^2 > 4ac$. This is the fundamental condition we need to satisfy for favorable outcomes.

Step 2: Determine the total number of possible outcomes (the sample space). The coefficients $a, b, c$ are chosen from the set $\{1, 2, 3, 4, 5, 6\}$. This means:

• There are 6 possible choices for $a$.
• There are 6 possible choices for $b$.
• There are 6 possible choices for $c$. Since the choices for $a, b,$ and $c$ are independent, the total number of unique combinations for the triplet $(a, b, c)$ is the product of the number of choices for each coefficient. $\text{Total number of outcomes} = 6 \times 6 \times 6 = 216$ This value represents the size of our sample space.

Step 3: Count the number of favorable outcomes. We need to find the number of triplets $(a, b, c)$ such that $a, b, c \in \{1, 2, 3, 4, 5, 6\}$ and $b^2 > 4ac$. We will systematically iterate through the possible values of $b$ and, for each $b$, determine the valid pairs of $(a,c)$.

• Case 1: $b = 1$ The condition $b^2 > 4ac$ becomes $1^2 > 4ac \implies 1 > 4ac$. Since $a, c \ge 1$, the minimum value of $4ac$ is $4 \times 1 \times 1 = 4$. As $1 > 4$ is false, there are 0 favorable outcomes for $b=1$.

• Case 2: $b = 2$ The condition $b^2 > 4ac$ becomes $2^2 > 4ac \implies 4 > 4ac \implies 1 > ac$. Since $a, c \ge 1$, the minimum value of $ac$ is $1 \times 1 = 1$. As $1 > ac$ is false for any $ac \ge 1$, there are 0 favorable outcomes for $b=2$.

• Case 3: $b = 3$ The condition $b^2 > 4ac$ becomes $3^2 > 4ac \implies 9 > 4ac \implies ac < \frac{9}{4} = 2.25$. Since $a, c$ are integers, possible values for $ac$ are $1$ and $2$.

  • If $ac = 1$: The only pair $(a,c)$ from $\{1, \dots, 6\}$ is $(1,1)$. (1 outcome)
  • If $ac = 2$: The pairs are $(1,2)$ and $(2,1)$. (2 outcomes) Total favorable outcomes for $b=3$: $1 + 2 = \mathbf{3}$.

• Case 4: $b = 4$ The condition $b^2 > 4ac$ becomes $4^2 > 4ac \implies 16 > 4ac \implies ac < 4$. Possible values for $ac$ are $1, 2, 3$.

  • If $ac = 1$: $(1,1)$. (1 outcome)
  • If $ac = 2$: $(1,2), (2,1)$. (2 outcomes)
  • If $ac = 3$: $(1,3), (3,1)$. (2 outcomes) Total favorable outcomes for $b=4$: $1 + 2 + 2 = \mathbf{5}$.

• Case 5: $b = 5$ The condition $b^2 > 4ac$ becomes $5^2 > 4ac \implies 25 > 4ac \implies ac < \frac{25}{4} = 6.25$. Possible values for $ac$ are $1, 2, 3, 4, 5, 6$.

  • If $ac = 1$: $(1,1)$. (1 outcome)
  • If $ac = 2$: $(1,2), (2,1)$. (2 outcomes)
  • If $ac = 3$: $(1,3), (3,1)$. (2 outcomes)
  • If $ac = 4$: $(1,4), (4,1), (2,2)$. (3 outcomes)
  • If $ac = 5$: $(1,5), (5,1)$. (2 outcomes)
  • If $ac = 6$: $(1,6), (6,1), (2,3), (3,2)$. (4 outcomes) Total favorable outcomes for $b=5$: $1 + 2 + 2 + 3 + 2 + 4 = \mathbf{14}$.

• Case 6: $b = 6$ The condition $b^2 > 4ac$ becomes $6^2 > 4ac \implies 36 > 4ac \implies ac < 9$. Possible values for $ac$ are $1, 2, 3, 4, 5, 6, 7, 8$. We must also ensure $a, c \in \{1, \dots, 6\}$.

  • If $ac = 1$: $(1,1)$. (1 outcome)
  • If $ac = 2$: $(1,2), (2,1)$. (2 outcomes)
  • If $ac = 3$: $(1,3), (3,1)$. (2 outcomes)
  • If $ac = 4$: $(1,4), (4,1), (2,2)$. (3 outcomes)
  • If $ac = 5$: $(1,5), (5,1)$. (2 outcomes)
  • If $ac = 6$: $(1,6), (6,1), (2,3), (3,2)$. (4 outcomes)
  • If $ac = 7$: No pairs $(a,c)$ from the set $\{1, \dots, 6\}$ multiply to 7. (0 outcomes)
  • If $ac = 8$: $(2,4), (4,2)$. (2 outcomes) Total favorable outcomes for $b=6$: $1 + 2 + 2 + 3 + 2 + 4 + 0 + 2 = \mathbf{16}$.

Summing the favorable outcomes from all cases: $\text{Total favorable outcomes} = 0 + 0 + 3 + 5 + 14 + 16 = 38$

Step 4: Calculate the probability $p$. The probability $p$ is the ratio of the total number of favorable outcomes to the total number of possible outcomes: $p = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{38}{216}$

Step 5: Calculate the value of $216p$. The problem asks for the value of $216p$: $216p = 216 \times \frac{38}{216} = 38$

3. Common Mistakes & Tips

• Misinterpretation of "one real root bigger than the other": A common error is to interpret this as "real roots" ($D \ge 0$), which would include cases where $D=0$ (repeated roots). However, "bigger than the other" explicitly implies distinctness, so $D>0$ is essential.
• Systematic Counting: When enumerating pairs $(a,c)$ for each $b$, ensure you are systematic. It's helpful to list them out to avoid missing any combinations. Remember that $(a,c)$ is an ordered pair, so $(1,2)$ is different from $(2,1)$ unless $a=c$.
• Checking Bounds: Always ensure that the values for $a$ and $c$ (and $b$) strictly adhere to the given set $\{1, 2, 3, 4, 5, 6\}$. For example, for $ac=7$, no valid pairs exist within the given set.

4. Summary

This problem required us to first translate the condition about the roots of a quadratic equation into an inequality involving its coefficients ($b^2 > 4ac$). We then determined the total number of possible combinations for the coefficients $a, b, c$ from the given set. The core of the solution involved systematically counting the number of favorable combinations by iterating through possible values of $b$ and finding corresponding pairs of $(a,c)$ that satisfy the inequality. Finally, we used the probability formula to find $p$ and then calculated $216p$.

The final answer is $\boxed{38}$, which corresponds to option (A).