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

We have $x^2-(a-5) x+\left(a-\frac{15}{4}\right)=0$
Now, discriminant $\geq 0$
$\Rightarrow(a-5)^2-4\left(a-\frac{15}{4}\right) \geq 0 \Rightarrow(a-4)(a-10) \geq 0 \Rightarrow a \in(-\infty, 4] \cup[10, \infty)$
CASE-I: When both roots are positive.
$D \geq 0, a-5>0, a-\frac{15}{4}>0 \Rightarrow D \geq 0, a>5, a \geq \frac{15}{4}, \text { so } a \in[10, \infty)$
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 $[1,100]=(1+2+3)+(10+11+12+\ldots \ldots .+100)$
$=6+5005=5011 $