If a, b, c are positive integers such that a>b>c and the quadratic equation (a + b–2c) x² +(b + c–2a) x+ (c + a–2b) = 0 has a root in the interval (–1,0) then

Clearly 1 is a root of the given equation .Given that 2nd root lies in (-1,0) Product of roots < 0 is
$\frac{\mathrm{c}+\mathrm{a}-2\mathrm{b}}{\mathrm{a}+\mathrm{b}-2\mathrm{c}}<0\Rightarrow \mathrm{c}+\mathrm{a}-\mathrm{zb}<0(\because \mathrm{a}+\mathrm{b}-2\mathrm{c}>0)$
The roots of the equation are both rational for the equation ax² + zbx + c = 0 we have f(0) = C>0
F(–1) = c + a - 2b < 0. hence one root is –ve 
Also for an equation with +ve real coefficients all roots are –ve hence 2 nd root is also –ve.