The coefficients a, b, c in the quadratic equation ax² + bx + c = 0 are from the set {1, 2, 3, 4, 5,

Q: Why $D > 0$ and not $D \ge 0$?

This is a fantastic question! The problem states that one root is strictly bigger than the other. If $D = 0$, the roots are equal (i.e., $x_1 = x_2$). If the roots are equal, you cannot say one is bigger than the other. Therefore, we must exclude the case where $D = 0$ and only consider $D > 0$ for distinct roots.

Q: Why iterate through $b$ and not $a$ or $c$?

Think about the inequality $b^2 > 4ac$. The variable $b$ is squared, which means it grows much faster than $a$ or $c$. By fixing $b$, we create a clear 'ceiling' for the product $ac$. It is the most efficient way to partition the problem into manageable sub-cases without missing any combinations.

Q: Is there a faster way to count than listing them manually?

While you could use symmetry or combinatorial arguments, for a small set like $\{1, 2, 3, 4, 5, 6\}$, manual counting is actually the safest and most robust method. It prevents 'over-counting' or 'under-counting' errors that often happen with complex formulas. In JEE Advanced, accuracy is just as important as speed!

Question

The coefficients $a, b, c$ in the quadratic equation $ax^2 + bx + c = 0$ are from the set $\{1, 2, 3, 4, 5, 6\}$. If the probability of this equation having one real root bigger than the other is $p$, then $216 p$ equals :

Options:

(A) 57

(B) 76

(C) 38

(D) 19

Sigma Percentile · Accepted Answer

### Analyzing the Setup The problem asks for the probability that the quadratic equation $ax^2 + bx + c = 0$ has one real root strictly larger than the other, given that $a, b, c \in \{1, 2, 3, 4, 5, 6\}$. For one root to be strictly larger than the other, the roots must be real and distinct. This condition is satisfied if and only if the discriminant $D$ is strictly greater than zero. $$D = b^2 - 4ac > 0 \quad \Rightarrow \quad b^2 > 4ac$$ ### The Sample Space We are choosing three coefficients $a, b,$ and $c$ from a set of 6 possible values. Since each choice is independent, the total number of possible equations is: $$6 imes 6 imes 6 = 216$$ This value represents our total sample space, which will serve as the denominator for our probability calculation. ### The Systematic Grind To find the number of favorable cases, we iterate through each possible value of $b$ and determine how many pairs $(a, c)$ satisfy the inequality $ac < \frac{b^2}{4}$. For $b=1$: $ac < 0.25$. Since the minimum value for $a$ and $c$ is 1, there are **0 cases**. For $b=2$: $ac < 1$. This is impossible, resulting in **0 cases**. For $b=3$: $ac < 2.25$. The valid pairs $(a, c)$ are $(1, 1), (1, 2),$ and $(2, 1)$. This gives **3 cases**. For $b=4$: $ac < 4$. The valid pairs are $(1, 1), (1, 2), (1, 3), (2, 1),$ and $(3, 1)$. This gives **5 cases**. For $b=5$: $ac < 6.25$. The valid pairs are $(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1), (5, 1),$ and $(6, 1)$. This gives **14 cases**. For $b=6$: $ac < 9$. The valid pairs are the 14 cases from $b=5$ plus $(2, 4)$ and $(4, 2)$. This gives **16 cases**. ### Final Calculation Summing the favorable cases across all values of $b$, we get: $$3 + 5 + 14 + 16 = 38$$ The probability $p$ is therefore: $$p = \frac{38}{216}$$ The problem asks for the value of $216p$. Multiplying our probability by the total sample space, we obtain: $$216 imes \left( \frac{38}{216} ight) = \mathbf{38}$$

Animated Solution for Mathematics - Probability: The coefficients $a, b, c$ in the quadratic equation $a x^{2} + b x + c = 0$ are from the set ${1, 2, 3, 4, 5, 6}$ . If the probability of this equation having one real root bigger than the other is $p$ , then $216 p$ equals :

Select Answer:

Visualized Solution

$a x^{2} + b x + c = 0$

Discriminant Condition

Total Sample Space

Analyzing $b = 1$ and $b = 2$

Case $b = 3$

Case $b = 4$

Case $b = 5$

Case $b = 6$

Total Favorable Cases

Final Probability Calculation

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Sample Space

The Systematic Grind

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ax2+bx+c=0

Discriminant Condition

Total Sample Space

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b=1 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b=2

Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b=3

Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b=4

Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b=5

Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b=6