Question

Find sum of all integral values of $a$ in $[1,100]$ for which the equation $x^2-(a-5)x+\left(a-\frac{15}{4}\right)=0$ has atleast one root greater than zero.

Answer 1

Answer: 5011

We have $x^2-(a-5)x+\left(a-\frac{15}{4}\right)=0$
Now, discriminant $\ge 0$
$\Rightarrow (a-5{)}^2-4\left(a-\frac{15}{4}\right)\ge 0\Rightarrow (a-4)(a-10)\ge 0\Rightarrow a\in (-\infty ,4]\cup [10,\infty )$
CASE-I: When both roots are positive.
$D\ge 0,a-5>0,a-\frac{15}{4}>0\Rightarrow D\ge 0,a>5,a\ge \frac{15}{4},\text{ so }a\in [10,\infty )$
CASE-II: When exactly one root is positive.
$a-\frac{15}{4}<0\text{, so }a<\frac{15}{4}$....(2)
$\therefore$ From (1) and (2), we conclude that $a\in \left(-\infty ,\frac{15}{4}\right)\cup [10,\infty )$
Hence, sum of all integral values of a in $[1,100]=(1+2+3)+(10+11+12+\dots \dots .+100)$
$=6+5005=5011$