If f(x) is a quadratic expression such that f(1)+f(2)=0, and -1 is a root of f(x) = 0,  

then the other root of f(x) =0 is

 

(A) -5/8

(B) -8/5

(C) 5/8

(D) 8/5

 

$$\begin{gathered} \mathrm{let} \mathrm{f}\left(\mathrm{x}\right)= {\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c} =0 \\ \mathrm{given} \mathrm{f}\left(1\right)+\mathrm{f}\left(2\right) =0 \& \mathrm{f}(-1)=0 \\ (\mathrm{a}+\mathrm{b}+\mathrm{c}) + 3\mathrm{a} +2\mathrm{b} +\mathrm{c} =0 \mathrm{a}{(-1)}^2-\mathrm{b}+\mathrm{c}=0 \\ = 5\mathrm{a} +3\mathrm{b} +2\mathrm{c} =0 \mathrm{a}-\mathrm{b}+\mathrm{c}=0 \\ \mathrm{using} \left(1\right) \& 11\times 43 \\ 5\mathrm{a}+\cancel{3\mathrm{b}}+2\mathrm{C}=0 \\ 3\mathrm{a}-\cancel{3\mathrm{b}}+3\mathrm{c}=0 \\ 8\mathrm{a}+ 5\mathrm{c}=0 \\ \mathrm{we} \mathrm{know} \mathrm{\alpha \beta } =4\mathrm{a} \mathrm{as} \mathrm{\alpha }=-1 \\ \mathbf{\beta }\mathbf{=}\frac{\mathbf{-}\mathbf{c}}{\mathbf{a}}\mathbf{=}\frac{\mathbf{8}}{\mathbf{5}}\mathbf{ }\mathbf{ }\mathbf{\left(}\mathbf{d}\mathbf{\right)} \end{gathered}$$