Probability of Quadratic Function Coefficients

For $f(x) = ax^2 + bx + a$ to have two irrational roots: Discriminant $D = b^2 - 4a^2 > 0$ and $D$ is not a perfect square. So $b^2 > 4a^2 \implies b > 2a$. Let's list all valid pairs $(a,b)$ with $a,b \in \{1,2,\dots,9\}$:

• $b=3: a=1 \implies D = 9 - 4 = 5$ (irrational) $\rightarrow (1,3)$, $a+b=4$
• $b=4: a=1 \implies D = 16 - 4 = 12$ (irrational) $\rightarrow (1,4)$, $a+b=5$
• $b=5: a=1 \implies D = 25 - 4 = 21$ (irrational) $\rightarrow (1,5)$, $a+b=6$
• $b=6: a=1 \implies D = 32$ (irrational), $a=2 \implies D = 36 - 16 = 20$ (irrational) $\rightarrow (1,6), (2,6)$
• $b=7: a=1,2,3 \implies D = 45, 33, 13$ (all irrational) $\rightarrow (1,7), (2,7), (3,7)$
• $b=8: a=1,2,3 \implies D = 60, 48, 28$ (all irrational) $\rightarrow (1,8), (2,8), (3,8)$
• $b=9: a=1,2,3,4 \implies D = 77, 65, 45, 17$ (all irrational) $\rightarrow (1,9), (2,9), (3,9), (4,9)$ Total permissible pairs = $1 + 1 + 1 + 2 + 3 + 3 + 4 = 15$. Pairs with $a+b > 9$:
• $b=6: (2,6) \implies 8 \le 9$ (No)
• $b=7: (3,7) \implies 10 > 9$ (1)
• $b=8: (2,8), (3,8) \implies 10, 11 > 9$ (2)
• $b=9: (1,9), (2,9), (3,9), (4,9) \implies 10, 11, 12, 13 > 9$ (4) Total pairs with $a+b > 9 = 1 + 2 + 4 = 7$. Probability = 7/15.