If the equation ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0(\mathrm{a}>0)$ has two roots $\mathrm{\alpha } \mathrm{and} \mathrm{\beta }$ such that $\mathrm{\alpha }<-2\text{ and }\mathrm{\beta }>2,$ then

Let y = ax² + bx + c
 

consider the following cases :
Case I D>0
$\Rightarrow {\mathrm{b}}^2-4\mathrm{ac}>0$
Case II af(-2)<0
→ a (4a -2b +c) <0
 → 4a-2b +c<0
Case III af(2)<0
→ a(4a + 2b+ c)<0
→ 4a+2b + c<0
Combining Case lI and Case III, we get
$4\mathrm{a}+2\left|\mathrm{b}\right|+\mathrm{c}<0$
Also, at x = 0, y <0$\Rightarrow$c<0
Also, since for -2 < x <2,
y <0
ax²+bx +c <0
For x= 1, a+b+ c <0 …..(i)
and for x= -1, a -b+ c <0…..(ii)
Combining Eqs. (i) and (ii), we get
$\mathrm{a}+\left|\mathrm{b}\right|+\mathrm{c}<0$