Answer: $19$

Explanation:1. Condition for no real solution of a quadratic equation: A quadratic equation $Ax^2+Bx+C=0$ has no real solution if its discriminant $D = B^2-4AC$ is less than zero ($D < 0$).

2. Identify the coefficients A, B, and C from the given equation: Given equation: $4x^2+bx+25=0$ - $A = 4$ - $B = b$ - $C = 25$

3. Set up the inequality for the discriminant: $D = b^2 - 4(4)(25) < 0$ $b^2 - 16 \cdot 25 < 0$ $b^2 - 400 < 0$

4. Solve the inequality for b: $b^2 < 400$ Taking the square root of both sides (and considering the implications for inequalities): If $b^2 < 400$, then $-\sqrt{400} < b < \sqrt{400}$. $-\sqrt{400} = -20$ $\sqrt{400} = 20$ So, $-20 < b < 20$.

5. Consider the condition that b is a positive integer: We have $-20 < b < 20$. Since b must be a positive integer, b can take values $1, 2, 3, \dots, 19$. (b cannot be 20, because if $b=20$, then $b^2=400$, and $b^2-400=0$, which means one real solution, not no real solution).

6. Determine the largest possible integer value of b: The positive integers b that satisfy $-20 < b < 20$ are $1, 2, \dots, 19$. The largest of these is $19$.

7. Verification: If $b=19$, $b^2 = 19^2 = 361$. Discriminant $D = 361 - 400 = -39$. Since $-39 < 0$, there are no real solutions. If $b=20$, $b^2 = 400$. Discriminant $D = 400 - 400 = 0$. This means one real solution. So, b must be strictly less than $20$.

8. Conclusion: The largest possible positive integer value of b for which the equation has no real solution is $19$.