Mathematics - Quadratic Equation Question with Solution | TestHub

Let $f(x)=a{x}^2+bx+c$. The expression is $Y=\frac{f(-1)}{f(1)}$. From $a+c>b>0$, we deduce $f(-1)=a-b+c>0$ and $f(1)=a+b+c>0$, hence $Y>0$. Let $\alpha ,\beta$ be the distinct real roots. Case 1: $a>0$. This implies $\alpha <\beta <-1$. Thus $1+\alpha ,1+\beta <0$ and $1-\alpha ,1-\beta >0$. So $Y=\frac{(1+\alpha )(1+\beta )}{(1-\alpha )(1-\beta )}\in (0,1)$. Case 2: $a<0$. This implies $\alpha <-1<1<\beta$. Thus $1+\alpha <0,1+\beta >0$ and $1-\alpha >0,1-\beta <0$. So $Y=\frac{(1+\alpha )(1+\beta )}{(1-\alpha )(1-\beta )}\in (0,\infty )$. Combining both cases, the range is $(0,\infty )$.