উদাহরণ ২৷ [tex]\alpha ,\beta [/tex] এবং [tex]\gamma ,\delta [/tex] যথাক্রমে [tex]{x^2} + px - r = 0[/tex] এবং [tex]{x^2} + px + r = 0[/tex] সমীকরণের বীজ হলে, প্রমান করো যে, [tex]\left( {\alpha - \gamma } \right)\left( {\alpha - \delta } \right) = \left( {\beta - \gamma } \right)\left( {\beta - \delta } \right)[/tex] . [Jt. Ent. ‘83 ]
সমাধান: [tex]\alpha ,\beta [/tex] এবং [tex]\gamma ,\delta [/tex] যথাক্রমে [tex]{x^2} + px - r = 0[/tex] এবং [tex]{x^2} + px + r = 0[/tex] সমীকরণের বীজ।
[tex]\begin{array}{l}
{\alpha ^2} + p\alpha - r = 0 \to \left( 1 \right)\\
{\beta ^2} + p\beta - r = 0 \to \left( 2 \right)\\
\gamma + \delta = - p \to \left( 3 \right)
\end{array}[/tex]
(1)- (2) করে পাই
[tex]\begin{array}{l}
{\alpha ^2} - {\beta ^2} + p\left( {\alpha - \beta } \right) = 0\\
\Rightarrow {\alpha ^2} - {\beta ^2} - \left( {\gamma + \delta } \right)\left( {\alpha - \beta } \right) = 0\left[ {\gamma + \delta = - p} \right]\\
\Rightarrow {\alpha ^2} - {\beta ^2} - \gamma \alpha + \gamma \beta - \delta \alpha + \delta \beta = 0\\
\Rightarrow \alpha \left( {\alpha - \gamma } \right) - \beta \left( {\beta - \gamma } \right) - \delta \alpha + \delta \beta = 0\\
\Rightarrow \alpha \left( {\alpha - \gamma } \right) - \beta \left( {\beta - \gamma } \right) - \delta \alpha + \delta \beta + \gamma \delta - \delta \gamma = 0\\
\Rightarrow \alpha \left( {\alpha - \gamma } \right) - \beta \left( {\beta - \gamma } \right) - \delta \left( {\alpha - \gamma } \right) + \delta \left( {\beta - \gamma } \right) = 0\\
\Rightarrow \left( {\alpha - \gamma } \right)\left( {\alpha - \delta } \right) - \left( {\beta - \gamma } \right)\left( {\beta - \delta } \right) = 0\\
\Rightarrow \left( {\alpha - \gamma } \right)\left( {\alpha - \delta } \right) = \left( {\beta - \gamma } \right)\left( {\beta - \delta } \right)\left[ {proved} \right]
\end{array}[/tex]