Probability of Strictly Positive Quadratic Expression

To find the probability that the quadratic expression $ax^2 + 2\sqrt{2}bx + c > 0$ for all $x \in \mathbb{R}$, where $a, b, c \in \{1, 2, 3, 4\}$, we first determine the condition for a quadratic expression to be strictly positive for all real values of $x$.

A quadratic expression $Ax^2 + Bx + C > 0$ for all $x \in \mathbb{R}$ if and only if:

1. $A > 0$
2. The discriminant $D = B^2 - 4AC < 0$

Here, $A = a$, $B = 2\sqrt{2}b$, and $C = c$. Since $a \in \{1, 2, 3, 4\}$, $a > 0$ is always satisfied.

Now, we calculate the discriminant $D$: $D = (2\sqrt{2}b)^2 - 4(a)(c) = 8b^2 - 4ac$

For $D < 0$, we require: $8b^2 - 4ac < 0 \implies ac > 2b^2$

The total number of possible ordered triplets $(a, b, c)$ chosen from $\{1, 2, 3, 4\}^3$ is: $\text{Total Outcomes} = 4 \times 4 \times 4 = 64$

Now, we count the number of favorable triplets $(a, b, c)$ satisfying $ac > 2b^2$ for each possible value of $b \in \{1, 2, 3, 4\}$:

1. Case 1: $b = 1$ $ac > 2(1)^2 = 2$ Out of the 16 total pairs $(a, c) \in \{1, 2, 3, 4\} \times \{1, 2, 3, 4\}$, the pairs where $ac \le 2$ are: $(1, 1), (1, 2), (2, 1)$ Thus, the number of favorable pairs for $b = 1$ is: $16 - 3 = 13$

2. Case 2: $b = 2$ $ac > 2(2)^2 = 8$ The pairs $(a, c)$ that satisfy $ac > 8$ are: $(3, 3), (3, 4), (4, 3), (4, 4)$ Thus, the number of favorable pairs for $b = 2$ is $4$.

3. Case 3: $b = 3$ $ac > 2(3)^2 = 18$ Since the maximum possible value of $ac$ is $4 \times 4 = 16$, no pairs satisfy $ac > 18$. Thus, the number of favorable pairs for $b = 3$ is $0$.

4. Case 4: $b = 4$ $ac > 2(4)^2 = 32$ Since the maximum possible value of $ac$ is $16$, no pairs satisfy $ac > 32$. Thus, the number of favorable pairs for $b = 4$ is $0$.

Summing the favorable outcomes from all cases gives: $\text{Total Favorable Outcomes} = 13 + 4 + 0 + 0 = 17$

The required probability is: $P = \frac{17}{64}$

Given that $P = \frac{m}{n}$ with $\gcd(m, n) = 1$, we have $m = 17$ and $n = 64$.

Finally, computing $m + n$: $m + n = 17 + 64 = 81$