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

Concept: For a quadratic expression \[ ax^2+bx+c>0 \quad \forall x\in\mathbb{R}, \] two conditions must hold:

• \(a>0\)
• Discriminant \(<0\)

The discriminant of \[ ax^2+2\sqrt2\,bx+c \] is \[ D=(2\sqrt2\,b)^2-4ac \] \[ =8b^2-4ac \] For positivity, \[ 8b^2-4ac<0 \] \[ 2b^2<ac \]
Step 1:Total possible choices. Since \[ a,b,c \in \{1,2,3,4\} \] Total outcomes: \[ 4^3=64 \]
Step 2:Check the inequality \(2b^2<ac\). For \(b=1\): \[ 2<ac \] Valid \((a,c)\): \[ (1,3),(1,4),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4) \] Total \(=13\). For \(b=2\): \[ 8<ac \] Valid pairs: \[ (3,3),(3,4),(4,3),(4,4) \] Total \(=4\). For \(b=3\): \[ 18<ac \] Possible pair: \[ (4,4) \] Total \(=1\). For \(b=4\): \[ 32<ac \] No solutions.
Step 3:Total favorable cases \[ 13+4+1=18 \] Probability: \[ \frac{18}{64}=\frac{9}{32} \] Thus \[ m=9,\quad n=32 \]
Step 4:Compute \(m+n\)} \[ 9+32=41 \] \[ \boxed{41} \]