Quadratic equation $x^2+[a^2-5a+b+4]x+b=0$ has roots $-5$ and $1$, then number of integral values of $a$ are Note: $[.]$ denotes the greatest integer function.

Using the given roots of -5 and 1 for the quadratic x^2 + [a^2 - 5a + b + 4]x + b = 0, the product of the roots gives b = (-5)(1) = -5. The sum of the roots yields -4 = -(a^2 - 5a + b + 4), which simplifies to the greatest integer of (a^2 - 5a - 1) being zero. This inequality dictates that a must lie strictly between 1 and 4, meaning the only possible integer values for a are 2 and 3. Testing both integers shows that a^2 - 5a - 1 evaluates to -7 and -9 respectively, which are both less than -1 and invalid, resulting in zero integral values for a.

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