উদাহরণ ৬৷ [tex]3{x^2} - 18x + 2 = 0[/tex] সমীকরণের বীজ দুটি [tex]\left( { - \frac{{{\alpha ^2}}}{\beta }} \right),\left( { - \frac{{{\beta ^2}}}{\alpha }} \right)[/tex] হলে, যে দ্বিঘাত সমীকরণর বীজ দুটি [tex]\alpha ,\beta [/tex]( [tex]\alpha ,\beta [/tex] বাস্তব) তা নির্ণয় করো। [H.S ‘89]
সমাধান: [tex]3{x^2} - 18x + 2 = 0[/tex] সমীকরণের বীজ দুটি হল [tex]\left( { - \frac{{{\alpha ^2}}}{\beta }} \right),\left( { - \frac{{{\beta ^2}}}{\alpha }} \right)[/tex]।
অতএব
[tex]\begin{array}{l}
\left( { - \frac{{{\alpha ^2}}}{\beta }} \right) + \left( { - \frac{{{\beta ^2}}}{\alpha }} \right) = \frac{{18}}{3} = 6 \to \left( 1 \right)\\
\left( { - \frac{{{\alpha ^2}}}{\beta }} \right) \times \left( { - \frac{{{\beta ^2}}}{\alpha }} \right) = \frac{2}{3} \to \left( 2 \right)
\end{array}[/tex]
(1) ও (2) থেকে পাই
[tex]\begin{array}{l}
\left( { - \frac{{{\alpha ^2}}}{\beta }} \right) \times \left( { - \frac{{{\beta ^2}}}{\alpha }} \right) = \frac{2}{3}\\
\Rightarrow \alpha \beta = \frac{2}{3} \to \left( 3 \right)\\
\left( { - \frac{{{\alpha ^2}}}{\beta }} \right) + \left( { - \frac{{{\beta ^2}}}{\alpha }} \right) = 6\\
\Rightarrow \frac{{{\alpha ^3} + {\beta ^3}}}{{\alpha \beta }} = - 6\\
\Rightarrow {\alpha ^3} + {\beta ^3} = - 6\alpha \beta \\
\Rightarrow {\left( {\alpha + \beta } \right)^3} - 3\alpha \beta \left( {\alpha + \beta } \right) = - 6\alpha \beta \\
\Rightarrow {\left( {\alpha + \beta } \right)^3} - 3 \times \frac{2}{3}\left( {\alpha + \beta } \right) = - 6 \times \frac{2}{3}\\
\Rightarrow {\left( {\alpha + \beta } \right)^3} - 2\left( {\alpha + \beta } \right) = - 4 \to \left( 4 \right)
\end{array}[/tex]
প্রশ্নানুযায়ী [tex]\alpha ,\beta [/tex] বাস্তব অতএব [tex]\left( {\alpha + \beta } \right)[/tex] এর মান বাস্তব হবে, এখন (4) নং সমীকরণে র বাস্তব সমাধান হয় [tex]\alpha + \beta = 2[/tex] এর জন্য।
সুতরাং দ্বিঘাত সমীকরণটি হবে
[tex]\begin{array}{l}
{x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta = 0\\
\Rightarrow {x^2} - 2x + \frac{2}{3} = 0\\
\Rightarrow 3{x^2} - 6x + 2 = 0
\end{array}[/tex]