If  $a,b,c$ are positive rational numbers such that $a>b>c,$  and the quadratic equation  $(a+b-2c)x^2+(b+c-2a)x + ( c+ a - 2 b ) = 0$  has a root in the interval  $(-1,0)$ then

 

$\because a>b>c \mathrm{.....}\left(1\right)$  and given equation is
$(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=\mathrm{0.....}\left(2\right)$
$\therefore$ Eqn. (2) has a root in the interval  $(-1,0)$
$\therefore f(-1)f\left(0\right)<0$
$\Rightarrow (2a-b-c)(c+a-2b)<0 .....\left(3\right)$
From eqn. (1)
$a>b\Rightarrow a-b>0$
$\&a>c\Rightarrow a-c>0$
$\therefore 2a-b-c>0 .....\left(4\right)$
From eqns. (3) & (4)
$c+a-2b<0$
$c+a<2b$
And discriminant of eqn. (2)
$\Rightarrow D=9{(b-c)}^2$
 Both roots of the given equation are rational
$\left(4\right)\alpha +\beta =\frac{-2b}{a}<0,\alpha \beta =\frac{c}{a}>0$
And discriminant is  $4b^2-4ac>0$
$\therefore$ Roots are negative and real