Find the no of digits in the sum of all integral values of a in $[1,100]$ for which following condition satisfies. Find the no of digits in the sum of all integral values of a in $[1,100]$ for which the equation $x^2-\left(a-5\right)x+\left(a-\dfrac{15}{4}\right)=0$ has atleast one root greater than zero.

My attempt is as follows:-

$$D\ge0$$
$$(a-5)^2-4\left(a-\dfrac{15}{4}\right)\ge0$$
$$a^2+25-10a-(4a-15)\ge0$$
$$a^2-14a+40\ge0$$
$$(a-4)(a-10)\ge0$$
$$a\in(-\infty,4]\cup[10,\infty)$$

Case $1$: Both roots are greater than $0$

$f(0)&gt;0, \dfrac{\alpha+\beta}{2}&gt;0$

$$a&gt;\dfrac{15}{4}$$
$$a-5&gt;0\implies a&gt;5$$

Hence intersection of all conditions would give $a\in[10,\infty)$

Case $2$: One root is greater than $0$ and one root is lesser than $0$

$$f(0)\le0$$
$$a\le\dfrac{15}{4}$$

Hence intersection of all conditions would give $a\in\left(-\infty,\dfrac{15}{4}\right]$

So all integral values of $a\in[1,100]=\{1,2,3,10,11\cdots\cdots100\}$, hence sum will be $6+\dfrac{91}{2}\cdot(10+100)=6+5005=6+5005=6+5005=5011$

So answer should be $4$ but actual answer is $7$. I tried to debug it a lot but didn't find any mistake. Please help me in this.
What you did was basically correct, but there are a few issues. First, one small point is that with your ranges, when a value is included on a boundary, you should use a square bracket instead of a round one. For example, with the result of checking the discriminant being non-negative, you should have stated $a\in(-\infty,4]\cup[10,\infty)$ instead of $a\in(-\infty,4)\cup(10,\infty)$. I see later you mostly assumed it excluded those values (but it should include them as the discriminant is $\ge 0$), so this seems to have been intentional on your part.

With your Case $1$, I'm not sure why repeated the calculations for the discriminant done earlier. Perhaps it was just a copy &amp; paste mistake?

Another minor point is that, technically, you should have Case $2$ be for one root greater than $0$ and one root being less than or equal to 0. Thus, you would consider $a \le \frac{15}{4}$. However, since only integral values of $a$ are involved, it doesn't make any difference to the end results in this case.

Next, your list of integral values of $a$ is stated correctly, but on the next line with the summation, you excluded $10$. As such, the actual sum would be $5001 + 10 = 5011$ instead.

Note the sum of the digits is $7$, so undoubtedly the question had intended to ask this instead of for the number of digits. Also, since the maximum sum has $4$ digits, there's no way the actual sum could have $7$ digits. In addition, to me, at least, asking for the number of digits in this type of problem doesn't make a lot of sense. Unless you make a significant mistake, it's fairly easy to see the result is $\ge 1000$, but even if all values from $1$ to $100$ were summed, the result would be $5050$, so it's fairly clear the answer would be $4$.
What you did was basically correct, but there are a few issues. First, one small point is that with your ranges, when a value is included on a boundary, you should use a square bracket instead of a round one. For example, with the result of checking the discriminant being non-negative, you should have stated $a\in(-\infty,4]\cup[10,\infty)$ instead of $a\in(-\infty,4)\cup(10,\infty)$. I see later you mostly assumed it excluded those values (but it should include them as the discriminant is $\ge 0$), so this seems to have been intentional on your part.

With your Case $1$, I'm not sure why repeated the calculations for the discriminant done earlier. Perhaps it was just a copy &amp; paste mistake?

Another minor point is that, technically, you should have Case $2$ be for one root greater than $0$ and one root being less than or equal to 0. Thus, you would consider $a \le \frac{15}{4}$. However, since only integral values of $a$ are involved, it doesn't make any difference to the end results in this case.

Next, your list of integral values of $a$ is stated correctly, but on the next line with the summation, you excluded $10$. As such, the actual sum would be $5001 + 10 = 5011$ instead.

Note the sum of the digits is $7$, so undoubtedly the question had intended to ask this instead of for the number of digits. Also, since the maximum sum has $4$ digits, there's no way the actual sum could have $7$ digits. In addition, to me, at least, asking for the number of digits in this type of problem doesn't make a lot of sense. Unless you make a significant mistake, it's fairly easy to see the result is $\ge 1000$, but even if all values from $1$ to $100$ were summed, the result would be $5050$, so it's fairly clear the answer would be $4$.