QJEEM24Mae820: JEE Main 2024 Mathematics - Class 12 JEEM24Mae820 Solution: Let and denote the outcome of three independent rolls of a fair …

A quadratic equation $ax^2 + bx + c = 0$ has real roots if and only if its discriminant is non-negative. The discriminant $\Delta$ of the quadratic equation is given by:

$\Delta = b^2 - 4ac$

For the quadratic equation to have all real roots, the discriminant must be non-negative:

$\Delta \geq 0$

That means:

$b^2 - 4ac \geq 0$

Given that $a, b, c$ are the outcomes of rolling a fair tetrahedral die, they can each be one of the numbers 1, 2, 3, or 4. Our task is to determine the probability that this condition holds.

We need to analyze the cases where $b^2 \geq 4ac$ .

Let’s consider all possible values for $a$ , $b$ , and $c$ , and count how many of them satisfy the condition. Since there are 4 choices for each of the variables, there are a total of $4 \times 4 \times 4 = 64$ possible combinations.

Now, we count the valid combinations where $b^2 \geq 4ac$ :

• For $a = 1$ : $b^2 \geq 4c$
• $b = 1: 1 \geq 4c \rightarrow \text{(Not possible since } c \ \text{must be } \geq 1 \text{ and not zero)}$
• $b = 2: 4 \geq 4c \rightarrow c \leq 1 \rightarrow c = 1$ (1 case)
• $b = 3: 9 \geq 4c \rightarrow c \leq 2 \rightarrow c = 1 \text{ or } 2$ (2 cases)
• $b = 4: 16 \geq 4c \rightarrow c \leq 4 \rightarrow c = 1, 2, 3, 4$ (4 cases)

Total for $a = 1 = 1 + 2 + 4 = 7$

• For $a = 2$ : $b^2 \geq 8c$
• $b = 1: 1 \geq 8c \rightarrow \text{(Not possible)}$
• $b = 2: 4 \geq 8c \rightarrow \text{(Not possible)}$
• $b = 3: 9 \geq 8c \rightarrow c \leq 1$ (1 case)
• $b = 4: 16 \geq 8c \rightarrow c \leq 2$ (2 cases)

Total for $a = 2 = 1 + 2 = 3$

• For $a = 3$ : $b^2 \geq 12c$
• $b = 1: 1 \geq 12c \rightarrow \text{(Not possible)}$
• $b = 2: 4 \geq 12c \rightarrow \text{(Not possible)}$
• $b = 3: 9 \geq 12c \rightarrow \text{(Not possible)}$
• $b = 4: 16 \geq 12c \rightarrow c \leq 1$ (1 case)

Total for $a = 3 = 1$

• For $a = 4$ : $b^2 \geq 16c$
• $b = 1: 1 \geq 16c \rightarrow \text{(Not possible)}$
• $b = 2: 4 \geq 16c \rightarrow \text{(Not possible)}$
• $b = 3: 9 \geq 16c \rightarrow \text{(Not possible)}$
• $b = 4: 16 \geq 16c \rightarrow c \leq 1$ (1 case)

Total for $a = 4 = 1$

Adding up all the valid cases:

$7 + 3 + 1 + 1 = 12$

The total number of valid combinations is 12 out of 64. Thus, the probability is:

$\frac{12}{64} = \frac{3}{16}$

The value of $\mathrm{m} = 3$ and $\mathrm{n} = 16$ . The sum $\mathrm{m} + \mathrm{n} = 3 + 16 = 19$ .

Hence, the answer is 19.