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 $0$
$f(0)>0, \dfrac{\alpha+\beta}{2}>0$
$$a>\dfrac{15}{4}$$ $$a-5>0\implies a>5$$
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.