Question

Let \(a,b,c \in \{1,2,3,4\}\). If the probability that \[ ax^2 + 2\sqrt{2}\,bx + c>0 \quad \text{for all } x \in \mathbb{R} \] is \( \frac{m}{n} \), where \(\gcd(m,n)=1\), then \(m+n\) is equal to _____.

Answer 1

Correct Answer: 41

Step 1: Understanding the Concept:
For a quadratic expression \(f(x) = Ax^2 + Bx + C\) to be strictly positive for all \(x\), the conditions are \(A>0\) and discriminant \(D<0\). Since \(a \in \{1, 2, 3, 4\}\), \(a>0\) is always true. We just need to check \(D<0\).
Step 2: Key Formula or Approach:
1. Discriminant \(D = (2\sqrt{2}b)^2 - 4ac<0\).
2. Total outcomes in the sample space: \(4 \times 4 \times 4 = 64\).
Step 3: Detailed Explanation:
The condition \(D<0\) implies:
\[ (2\sqrt{2}b)^2 - 4ac<0 \implies 8b^2<4ac \implies 2b^2<ac \]
We check valid triplets \((a, b, c)\) by testing cases for \(b\):
Case 1: \(b = 1\)
\(2(1)^2<ac \implies ac>2\).
Total pairs \((a, c)\) are \(4 \times 4 = 16\).
Pairs where \(ac \leq 2\) are \((1, 1), (1, 2), (2, 1)\) (3 pairs).
Favorable pairs for \(b=1\) is \(16 - 3 = 13\).
Case 2: \(b = 2\)
\(2(2)^2<ac \implies ac>8\).
Valid \((a, c)\) pairs: \((3, 3), (3, 4), (4, 3), (4, 4)\) (4 pairs).
Case 3: \(b = 3\)
\(2(3)^2<ac \implies ac>18\).
Since the maximum value of \(ac\) is \(4 \times 4 = 16\), no pairs exist.
Case 4: \(b = 4\)
\(2(4)^2<ac \implies ac>32\). No pairs exist.
Total favorable outcomes = \(13 + 4 = 17\).
Probability \(P = \frac{17}{64} = \frac{m}{n}\).
Since \(\text{gcd}(17, 64) = 1\), \(m = 17\) and \(n = 64\).
Calculating \(m + n = 17 + 64 = 81\).
Step 4: Final Answer:
The value of \(m+n\) is 81.