Quadratic Function Irrational Roots Probability

For $f(x) = ax^2 + bx + a$ to have two irrational roots, discriminant $D = b^2 - 4a^2 > 0$ and $D$ must not be a perfect square. $b^2 > 4a^2 \implies b > 2a$.

Valid pairs $(a, b)$ with $a, b \in \{1, 2, \dots, 9\}$:

• $b = 3$: $a=1 \implies D = 9 - 4 = 5$ (irrational) $\implies (1,3)$
• $b = 4$: $a=1 \implies D = 16 - 4 = 12$ (irrational) $\implies (1,4)$
• $b = 5$: $a=1, 2 \implies$ for $a=2, D = 25 - 16 = 9$ (square, rational). $a=1 \implies (1,5)$
• $b = 6$: $a=1, 2 \implies D(1)=32, D(2)=20$ $\implies (1,6), (2,6)$
• $b = 7$: $a=1, 2, 3 \implies D(1)=45, D(2)=33, D(3)=13 \implies (1,7), (2,7), (3,7)$
• $b = 8$: $a=1, 2, 3 \implies D(1)=60, D(2)=48, D(3)=28 \implies (1,8), (2,8), (3,8)$
• $b = 9$: $a=1, 2, 3, 4 \implies D(1)=77, D(2)=65, D(3)=45, D(4)=17 \implies (1,9), (2,9), (3,9), (4,9)$

Total permissible pairs = $1 + 1 + 1 + 2 + 3 + 3 + 4 = 15$. Pairs with $a + b > 9$: $(1,9), (2,9), (3,9), (4,9), (2,8), (3,8), (3,7)$ $\implies 7$ pairs. Probability = $7/15$.