Given that ax^​2 + bx + c has always the same sign as c if

a) b² >4ac   b) c>0  c) c<0   d) none of these 

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c} \\ \mathrm{Case} \left(\mathrm{i}\right) \mathrm{c}>0 \\ \mathrm{If} \mathrm{c}>0, \mathrm{then} \mathrm{we} \mathrm{want} \\ {\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}>0 \mathrm{for} \mathrm{all} \mathrm{x}\in \mathrm{R} \\ \mathrm{which} \mathrm{is} \mathrm{possible} \mathrm{when} \mathrm{opening} \mathrm{is} \mathrm{upward} \left(\mathrm{a}>0\right) \mathrm{and} \mathrm{discrimimant} \mathrm{is} \mathrm{negative} \\ \mathrm{i}.\mathrm{e}. \mathrm{graph} \mathrm{does} \mathrm{not} \mathrm{cut} \mathrm{x}-\mathrm{axis}. \\ \mathrm{So} \mathrm{we} \mathrm{have} \\ \mathbf{a}\mathbf{>}\mathbf{0} \\ \mathbf{and} \\ {\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{ac}\mathbf{<}\mathbf{0} \\ \mathrm{Case} \left(\mathrm{ii}\right) \\ \mathrm{If} \mathrm{c}<0, \mathrm{we} \mathrm{want} \\ {\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}<0 \\ \mathrm{which} \mathrm{is} \mathrm{possible} \mathrm{when} \mathrm{opening} \mathrm{is} \mathrm{downward} \left(\mathrm{a}<0\right) \mathrm{and} \mathrm{discriminant} \mathrm{is} \mathrm{negative} \\ \mathrm{i}.\mathrm{e}. \mathrm{graph} \mathrm{does} \mathrm{not} \mathrm{cut} \mathrm{x}-\mathrm{axis}. \\ \mathbf{a}\mathbf{<}\mathbf{0} \\ \mathbf{and} \\ {\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{ac}\mathbf{<}\mathbf{0} \\ \mathrm{Hence} \mathrm{option} \left(\mathrm{d}\right) \mathrm{is} \mathrm{correct}. \end{gathered}$$

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