Problem 002 | Quadratic Equations

উদাহরণ ২৷ $\alpha ,\beta$ এবং $\gamma ,\delta$  যথাক্রমে ${x^2} + px - r = 0$ এবং ${x^2} + px + r = 0$  সমীকরণের বীজ হলে, প্রমান করো যে,     $\left( {\alpha  - \gamma } \right)\left( {\alpha  - \delta } \right) = \left( {\beta  - \gamma } \right)\left( {\beta  - \delta } \right)$   .                                [Jt. Ent. ‘83 ]

সমাধান: $\alpha ,\beta$ এবং $\gamma ,\delta$  যথাক্রমে ${x^2} + px - r = 0$ এবং  ${x^2} + px + r = 0$ সমীকরণের বীজ।

$\begin{array}{l} {\alpha ^2} + p\alpha  - r = 0 \to \left( 1 \right)\\ {\beta ^2} + p\beta  - r = 0 \to \left( 2 \right)\\ \gamma  + \delta  =  - p \to \left( 3 \right) \end{array}$

(1)- (2) করে পাই

$\begin{array}{l} {\alpha ^2} - {\beta ^2} + p\left( {\alpha  - \beta } \right) = 0\\  \Rightarrow {\alpha ^2} - {\beta ^2} - \left( {\gamma  + \delta } \right)\left( {\alpha  - \beta } \right) = 0\left[ {\gamma  + \delta  =  - p} \right]\\  \Rightarrow {\alpha ^2} - {\beta ^2} - \gamma \alpha  + \gamma \beta  - \delta \alpha  + \delta \beta  = 0\\  \Rightarrow \alpha \left( {\alpha  - \gamma } \right) - \beta \left( {\beta  - \gamma } \right) - \delta \alpha  + \delta \beta  = 0\\  \Rightarrow \alpha \left( {\alpha  - \gamma } \right) - \beta \left( {\beta  - \gamma } \right) - \delta \alpha  + \delta \beta  + \gamma \delta  - \delta \gamma  = 0\\  \Rightarrow \alpha \left( {\alpha  - \gamma } \right) - \beta \left( {\beta  - \gamma } \right) - \delta \left( {\alpha  - \gamma } \right) + \delta \left( {\beta  - \gamma } \right) = 0\\  \Rightarrow \left( {\alpha  - \gamma } \right)\left( {\alpha  - \delta } \right) - \left( {\beta  - \gamma } \right)\left( {\beta  - \delta } \right) = 0\\  \Rightarrow \left( {\alpha  - \gamma } \right)\left( {\alpha  - \delta } \right) = \left( {\beta  - \gamma } \right)\left( {\beta  - \delta } \right)\left[ {proved} \right] \end{array}$