সৃজনশীলঃ ‘5×2−7x+3=0 সমীকরণের মূলদ্বয়…’ ২০২৫ – বহুপদী ও বহুপদী সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $5x^2-7x+3=0$ সমীকরণের মূলদ্বয়  $\alpha ,\beta$ 

$\therefore \alpha +\beta =-\frac{-7}{5}=\frac{7}{5}$ এবং $\alpha \beta =\frac{3}{5}$ 

নির্ণেয় সমীকরণের মূলদ্বয়  ${\alpha }^3$ ও ${\beta }^3$ 

$\therefore {\alpha }^3+{\beta }^3=(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )$ 

$={\left(\frac{7}{5}\right)}^3-3\times \frac{3}{5}\times \frac{7}{5}=\frac{343}{125}-\frac{63}{25}=\frac{343-315}{125}=\frac{28}{125}$ 

এবং ${\alpha }^3\times {\beta }^3=(\alpha \beta {)}^3={\left(\frac{3}{5}\right)}^3=\frac{27}{125}$ ,

$\therefore {\alpha }^3$ ও ${\beta }^3$ মূলবিশিষ্ট সমীকরণ 

$x^2-({\alpha }^3+{\dot{\beta }}^3)x+{\alpha }^3\times {\beta }^3=0$ 

বা, $x^2-\frac{28}{125}x+\frac{27}{125}=0$ 

$\therefore 125x^2-28x+27=0$ 

নির্ণেয় সমীকরণ , $125x^2-28x+27=0$. 

দেওয়া আছে, $5x^2-7x+3=0$ সমীকরণের মূলদ্বয়  $\alpha ,\beta$. 

$\therefore \alpha +\beta =-\frac{-7}{5}=\frac{7}{5}$ এবং  $\alpha \beta =\frac{3}{5}$ 

প্রদত্ত রাশি $=(1+\alpha +{\alpha }^2)(1+\beta +{\beta }^2)$ 

$\begin{array}{cc}& =1+\beta +{\beta }^2+\alpha +\alpha {\beta }^1+\alpha {\beta }^2+{\alpha }^2+{\alpha }^2\beta +{\alpha }^2{\beta }^2\\ & =1+(\alpha +\beta )+\left({\alpha }^2\right)+{\beta }^2)+\alpha \beta (\alpha +\beta +1)+{\alpha }^2{\beta }^2\\ & =1+(\alpha +\beta )+(\alpha +\beta {)}^2+2\alpha \beta +\alpha \beta (\alpha +\beta +1)+(\alpha \beta {)}^2\end{array}$ 

$=1+\frac{7}{5}+{\left(\frac{7}{5}\right)}^2-2\times \frac{3}{5}+\frac{3}{5}(\frac{7}{5}+1)+{\left(\frac{3}{5}\right)}^2$

$\begin{array}{cc}& =1+\frac{7}{5}+\frac{49}{25}-\frac{6}{5}+\frac{3}{5}\times \frac{12}{5}+\frac{9}{25}\\ & =1+\frac{7}{5}+\frac{49}{25}-\frac{6}{5}+\frac{36}{25}+\frac{9}{25}\\ & =\frac{25+35+49-30+36+9}{25}=\frac{154-30}{25}=\frac{124}{25}\end{array}$

নির্ণেয় মান  : $\frac{124}{25}$