If probability that $ax^2 + 2\sqrt{2}bx + c > 0 \, \forall x \in R$ where $a, b, c \in {1, 2, 3, 4}$ is $\frac{m}{n}$ where $m$ & $n$ are coprime, then $(m + n)$ is:
Step 1: Understanding the Concept:
For a quadratic expression \( ax^2 + px + q>0 \) for all \( x \), we must have \( a>0 \) and the discriminant \( D<0 \). Step 2: Key Formula or Approach:
The condition \( D<0 \) gives:
\( (2\sqrt{2}b)^2 - 4ac<0 \implies 8b^2<4ac \implies 2b^2<ac \).
Total possible combinations of \( (a, b, c) \) are \( 4 \times 4 \times 4 = 64 \). Step 3: Detailed Explanation:
Let's check combinations based on values of \( b \): If \( b = 1 \): \( 2<ac \). Possible \( ac \) values: \( \{3, 4, 6, 8, 9, 12, 16\} \).
Pairs \( (a, c) \): (1,3), (1,4), (3,1), (4,1), (2,2), (2,3), (3,2), (2,4), (4,2), (3,3), (3,4), (4,3), (4,4). (Total = 13). If \( b = 2 \): \( 8<ac \). Possible \( ac \) values: \( \{9, 12, 16\} \).
Pairs \( (a, c) \): (3,3), (3,4), (4,3), (4,4). (Total = 4). If \( b = 3 \): \( 18<ac \). Max \( ac = 16 \), so 0 cases. If \( b = 4 \): \( 32<ac \). 0 cases.
Total favorable cases = \( 13 + 4 = 17 \).
Probability = \( \frac{17}{64} \). Since 17 and 64 are coprime, \( m = 17, n = 64 \).
\( m + n = 17 + 64 = 81 \). Step 4: Final Answer:
The sum \( m + n \) is 81.
