Let $$a, b, c$$  be real numbers, $$a \ne 0.$$  If $$\alpha $$ is a root of $${a^2}{x^2} + bx + c = 0.\,\,\beta $$     is the root of $${a^2}{x^2} - bx - c = 0$$    and $$0 < \alpha < \beta ,$$   then the equation $${a^2}{x^2} + 2bx + 2c = 0$$     has a root $$\gamma $$ that always satisfies

Solution :
KEY CONCEPT:
If $$f\left( \alpha \right)$$ and $$f\left( \beta \right)$$ are of opposite signs then there must lie a value $$\gamma $$ between $$\alpha $$ and $$\beta $$ such that $$f\left( \gamma \right) = 0.$$
$$a,b,c$$   are real numbers and $$a \ne 0.$$
$$\eqalign{ & {\text{As }}\,\alpha \,\,{\text{is a root of }}\,{a^2}{x^2} + bx + c = 0 \cr & \therefore \,\,{a^2}{\alpha ^2} + b\alpha + c = 0\,\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{Also }}\beta \,{\text{ is root of }}\,{a^2}{x^2} - bx - c = 0 \cr & \therefore \,\,{a^2}{\beta ^2} - b\beta - c = 0\,\,\,\,\,\,\,\,.....\left( 2 \right) \cr & {\text{Now, let}}\,\,f\left( x \right) = {a^2}{x^2} + 2bx + 2c \cr & {\text{Then}}\,\,f\left( \alpha \right) = {a^2}{\alpha ^2} + 2b\alpha + 2c \cr & = {a^2}{\alpha ^2} + 2\left( {b\alpha + c} \right) \cr & = {a^2}{\alpha ^2} + 2\left( { - {a^2}{\alpha ^2}} \right)\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{Using eq}}{\text{.}}\left( 1 \right)} \right] \cr & = - {a^2}{\alpha ^2}. \cr & {\text{and }}\,f\left( \beta \right) = {a^2}{\beta ^2} + 2b\beta + 2c \cr & = {a^2}{\beta ^2} + 2\left( {b\beta + c} \right) \cr & = {a^2}{\beta ^2} + 2\left( {{a^2}{\beta ^2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{Using eq}}{\text{.}}\left( 2 \right)} \right] \cr & = 3{a^2}{\beta ^2} > 0. \cr & {\text{Since }}\,f\left( \alpha \right)\,{\text{and }}f\left( \beta \right)\,{\text{are of opposite signs and }}\gamma {\text{ is a root}} \cr & {\text{of equation }}\,f\left( x \right) = 0 \cr & \therefore \,\,\gamma \,\,{\text{must lie between }}\alpha \,\,{\text{and}}\,\,\beta \cr & {\text{Thus }}\alpha < \gamma < \beta . \cr & \therefore \,\,\,\left( {\text{D}} \right){\text{is the correct option}}{\text{.}} \cr} $$