Sum Reciprocal Two

Note: The hidden concept used in this question is $t+\cfrac { 1 }{ t } \quad \ge \quad 2\quad \quad (if\quad t>0\quad )$ . (By AM-GM)

Let roots of quadratic as ${ r }_{ 1 }\quad ,\quad { r }_{ 2 }$ .

Clearly By AM-GM we conclude that one root is greater or equal to 2 and another is Less than or equal to 0 .

Let ${ r }_{ 1 }\quad >\quad { r }_{ 2 }$ .

$\displaystyle{ r }_{ 1 }=\quad \alpha +\cfrac { 1 }{ \alpha } \quad \ge \quad 2\\ { r }_{ 2 }=\quad 2-(\beta +\cfrac { 1 }{ \beta } )\quad \le \quad 0$ .

${ r }_{ 1 }\quad \ge \quad 2\quad \& \quad { r }_{ 2 }\quad \le \quad 0$ .

So condition's are :

$\displaystyle\bullet \quad f\left( 2 \right) \quad \le \quad 0\\ \bullet \quad f\left( 0 \right) \quad \le \quad 0\\ \bullet \quad D\quad >\quad 0\quad \quad \quad (Useless)$ .

By solving these condition's and taking intersection we should get:

$a\quad \in \quad [-1,3]$ .

whatever $\alpha$ the root $\alpha + \alpha^{-1}$ is greater or equal 2. You can use AM-GM to get this quick. For same reason the other root is lower equal $0$ . So we have as condition the product of two roots is negative or zero. Since $a - 3$ is the product of these roots it implies that w have as condition $a \leq 3$ Now the positive root is $(a+1) + \sqrt{(a +1)^2 -a + 3 }$ the fact that it has to be greater or equal 2 brings to the disequation

$(a+1) + \sqrt{(a +1)^2 -a + 3 } \geq 2$

that can be easily solved and leads to $a \geq -1$ .

Both conditions on the roots means that $a \in [-1 ; 3]$ So the sum of the integer values possible for $a$ is $S = -1 + 0 + 1 + 2 + 3 = 5$ .