The coefficients a, b and c of the quadratic equation, ax² + bx + c = 0 are obtained by throwing a d

Q: Wait, why does $b^2 = 4ac$ imply $b$ must be even?

Because $4ac$ is clearly a multiple of 4, the left side $b^2$ must also be a multiple of 4. If $b^2$ is divisible by 4, then $b$ itself must be even. If $b$ were odd, $b^2$ would be odd, which cannot equal a multiple of 4.

Q: Can we just list all 216 possibilities?

You could, but it's inefficient and prone to error. The parity argument is the 'JEE way'—it's elegant, saves time, and demonstrates a deeper understanding of the constraints.

Q: What if $a$ or $c$ were zero?

The problem states $a, b, c$ are outcomes of a die, so they are restricted to the set $\{1, 2, 3, 4, 5, 6\}$. They cannot be zero. Always check your domain before starting!

Question

The coefficients $a, b$ and $c$ of the quadratic equation, $ax^2 + bx + c = 0$ are obtained by throwing a dice three times. The probability that this equation has equal roots is :

Options:

(A) $\frac{1}{54}$

(B) $\frac{1}{72}$

(C) $\frac{1}{36}$

(D) $\frac{5}{216}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are holding a standard six-sided die. You roll it three times, and with each roll, you are defining the coefficients of a quadratic equation: $ax^2 + bx + c = 0$.

This is the beauty of probability—it turns the chaotic, random act of rolling dice into a precise, deterministic mathematical challenge. Our goal is to find the probability that this equation has equal roots.

### The Sample Space

Every probability problem begins with the sample space, the set of all possible outcomes. Since each coefficient $a, b,$ and $c$ is chosen from the set $\{1, 2, 3, 4, 5, 6\}$, and we have three independent rolls, the total number of combinations is:

$$6 	imes 6 	imes 6 = 216$$

This is our denominator, the foundation upon which we build our probability.

### The Discriminant Constraint

Now, we turn to the heart of the quadratic equation: the discriminant $D = b^2 - 4ac$. For the roots to be equal, the parabola must just 'kiss' the x-axis, which happens if and only if $D = 0$.

This gives us our golden constraint:

$$b^2 = 4ac$$

This equation is the key to the entire problem. It is not just an algebraic expression; it is a filter that separates the successful outcomes from the failures.

### The Parity Shortcut

Look closely at $b^2 = 4ac$. The right side is a multiple of 4, which means $b^2$ must also be a multiple of 4.

This forces $b$ to be an even number. Since $b$ is a die roll, $b$ can only be $2, 4,$ or $6$. This realization is a massive shortcut that allows us to bypass checking all 216 possibilities.

### Systematic Case Analysis

Let us test our cases individually:

**Case 1:** If $b = 2$, then $b^2 = 4$. Our equation becomes $4 = 4ac$, which simplifies to $ac = 1$. The only pair $(a, c)$ that satisfies this is $(1, 1)$. This yields **1** favorable outcome.

**Case 2:** If $b = 4$, then $b^2 = 16$. Our equation becomes $16 = 4ac$, or $ac = 4$. The pairs $(a, c)$ that multiply to 4 are $(1, 4), (4, 1),$ and $(2, 2)$. This yields **3** favorable outcomes.

**Case 3:** If $b = 6$, then $b^2 = 36$. Our equation becomes $36 = 4ac$, or $ac = 9$. The only pair $(a, c)$ that works is $(3, 3)$. This yields **1** favorable outcome.

### Final Calculation

We have systematically counted our successes: $1 + 3 + 1 = 5$. These are the only 5 scenarios out of 216 where our quadratic equation will have equal roots.

The final probability is the ratio of favorable outcomes to the total sample space:

$$P = \frac{5}{216}$$

It is a simple, elegant result born from a rigorous, logical process. You have just mastered the art of constrained counting.

Animated Solution for Mathematics - Probability: The coefficients $a, b$ and $c$ of the quadratic equation, $a x^{2} + b x + c = 0$ are obtained by throwing a dice three times. The probability that this equation has equal roots is :

Select Answer:

Visualized Solution

The Setup

Total Outcomes

Condition for Equal Roots

Analyzing $b^{2} = 4 a c$

Case 1: $b = 2$

Case 2: $b = 4$

Case 3: $b = 6$

Total Favorable Outcomes

Final Probability

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Sample Space

The Discriminant Constraint

The Parity Shortcut

Systematic Case Analysis

Final Calculation

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