সৃজনশীলঃ ‘দৃশ্যকল্প: x3+qx+r=0 সমীকরণের…’ যশোর বোর্ড ২০২৫ – বহুপদী ও বহুপদী সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $x^3+\mathrm{qx}+r=0$ 

বা, $x^3+0\cdot x^2+qx+r=0$$..........\left(1\right)$

$\left(1\right)$ নং সমীকরণে তিনটি মূল $\alpha ,\beta ,\gamma$ 

$\therefore \alpha +\beta +\gamma =0$ বা, $\beta +\gamma =-\alpha$ 

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{g}{1}=q$ 

এবং $\alpha \beta \gamma =-\frac{r}{1}=-r$ 

আবার, $\left(i\right)$ নং সমীকরণের একটি মূল $\alpha$ হওয়ায়, 

${\alpha }^3+q\alpha +r=0$ 

বা, ${\alpha }^3=-r-q\alpha$ 

বামপক্ষ $=(\beta -\gamma {)}^2=\left(\right(\beta +\gamma {)}^2-4\beta \gamma$ 

$=(-\alpha {)}^2-4\left(\frac{-r}{\alpha }\right)[\because \beta +\gamma =-\alpha$ এবং $\alpha \beta \gamma =-r]$ 

$={\alpha }^2+\frac{4r}{\alpha }=\frac{{\alpha }^3+4r}{\alpha }=\frac{-r-q\alpha +4r}{\alpha }=\frac{3r-q\alpha }{\alpha }=$ ডানপক্ষ 

$\therefore (\beta -\gamma {)}^2=\frac{3r-q\alpha }{\alpha }$. (দেখানো হলো ) 

দেওয়া আছে, $x^3+\mathrm{qx}+r=0$ 

বা, $x^3+0\cdot x^2+qx+r=0$$.......\left(i\right)$

$\left(1\right)$ নং সমীকরণের তিনটি মূল $\alpha ,\beta ,\gamma$ 

$\therefore \alpha +\beta +\gamma =0$ 

বা, $\alpha +\beta =-\gamma$
বা, $\beta +\gamma =-\alpha$
বা, $\alpha +\gamma =-\beta$

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{q}{1}=q$ 

এবং $\alpha \beta \gamma =-\frac{r}{1}=-r$ 

$\frac{\beta +\gamma }{{\alpha }^3},\frac{\gamma +\alpha }{{\beta }^3},\frac{\alpha +\beta }{{\gamma }^3}$  মূলবিশিষ্ট সমীকরণ নির্ণয় করতে হবে। 

এখন, 

$\frac{\beta +\gamma }{{\alpha }^3}+\frac{\gamma +\alpha }{{\beta }^3}+\frac{\alpha +\beta }{{\gamma }^3}$ 

$\begin{array}{cc}& =\frac{-\alpha }{{\alpha }^3}+\frac{-\beta }{{\beta }^3}+\frac{-\gamma }{{\gamma }^3}=\frac{-1}{{\alpha }^2}-\frac{1}{{\beta }^2}-\frac{1}{{\gamma }^2}=\frac{-{\beta }^2{\gamma }^2-{\gamma }^2{\alpha }^2-{\alpha }^2{\beta }^2}{{\alpha }^2{\beta }^2{\gamma }^2}\\ & =\frac{-({\alpha }^2{\beta }^2+{\beta }^2{\gamma }^2+{\gamma }^2{\alpha }^2)}{(\alpha \beta \gamma {)}^2}=\frac{-\left\{\right(\alpha \beta {)}^2+(\gamma \alpha {)}^2+(\beta \gamma {)}^2\}}{(\alpha \beta \gamma {)}^2}\\ & =\frac{-\left\{\right(\alpha \beta +\beta \gamma +\gamma \alpha {)}^2-2(\alpha \beta \cdot \beta \gamma +\beta \gamma \cdot \gamma \alpha +\gamma \alpha \cdot \alpha \beta )\}}{(\alpha \beta \gamma {)}^2}\\ & =\frac{-\left\{\right(\alpha \beta +\beta \gamma +\gamma \alpha {)}^2-2({\alpha }^2\beta \gamma +\alpha {\beta }^2\gamma +\alpha \beta {\gamma }^2)}{(\alpha \beta \gamma {)}^2}\\ & =\frac{-\left\{\right(\alpha \beta +\beta \gamma +\gamma \alpha {)}^2-2\alpha \beta \gamma (\alpha +\beta +\gamma )\}}{(\alpha \beta \gamma {)}^2}\\ & =\frac{-\{q^2-2(-r)\cdot 0\}}{(-r{)}^2}=\frac{-q^2}{r^2}\end{array}$ 

আবার, $\frac{\beta +\gamma }{{\alpha }^3}\cdot \frac{\gamma +\alpha }{{\beta }^3}+\frac{\gamma +\alpha }{{\beta }^3}\cdot \frac{\gamma +\beta }{{\gamma }^3}+\frac{\alpha +\beta }{{\gamma }^3}\cdot \frac{\beta +\gamma }{{\alpha }^3}$ 

$\begin{array}{cc}& =\frac{-\alpha }{{\alpha }^3}\cdot \frac{-\beta }{{\beta }^3}+\frac{-\beta }{{\beta }^3}\cdot \frac{-\alpha }{{\gamma }^3}+\frac{-\gamma }{{\gamma }^3}\cdot \frac{-\alpha }{{\alpha }^3}\\ & =\frac{1}{{\alpha }^2{\beta }^2}+\frac{1}{{\beta }^2{\gamma }^2}+\frac{1}{{\alpha }^2{\gamma }^2}=\frac{{\gamma }^2+{\alpha }^2+{\beta }^2}{{\alpha }^2{\beta }^2{\gamma }^2}\\ & =\frac{(\alpha +\beta +\gamma {)}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )}{(\alpha \beta \gamma {)}^2}=\frac{0^2-2a}{(-r{)}^2}=\frac{-2a}{r^2}\end{array}$ 

আবার, $\frac{\beta +\gamma }{{\alpha }^3}\cdot \frac{\gamma +\alpha }{{\beta }^3}\cdot \frac{\alpha +\beta }{{\gamma }^3}=\frac{-\alpha }{{\alpha }^3}\cdot \frac{\mp \beta }{{\beta }^3}\cdot \frac{-\gamma }{{\gamma }^3}=\frac{-1}{{\alpha }^2{\beta }^2{\gamma }^2}=\frac{-1}{(\alpha \beta \gamma {)}^2}=\frac{-1}{(-r{)}^2}=\frac{-1}{r}$ নির্ণেয় $\frac{\beta +\gamma }{{\alpha }^3}\cdot \frac{\gamma +\alpha }{{\beta }^3}+\frac{\alpha +\beta }{{\gamma }^3}$ মূলবিশিষ্ট সমীকরণ, 

$x^2-(\frac{\beta +\gamma }{{\alpha }^3}+\frac{\gamma +\alpha }{{\beta }^3}+\frac{\alpha +\beta }{{\gamma }^3})x^2+(\frac{\beta +\gamma }{{\alpha }^3}\cdot \frac{\gamma +\alpha }{{\beta }^3}+\frac{\gamma +\alpha }{{\beta }^3}\cdot \frac{\alpha +\beta }{{\gamma }^3}+\frac{\alpha +\beta }{\gamma }\cdot \frac{\beta +\gamma }{{\alpha }^3})x$ 

$-(\frac{\beta +\gamma }{{\alpha }^3}\cdot \frac{\gamma +\alpha }{{\beta }^3}\cdot \frac{\alpha +\beta }{{\gamma }^3})=0$ 

$x^3-(-\frac{q^2}{r^2})x^2+(-\frac{2q}{r^2})x-(-\frac{1}{r^2})=0$ 

বা, $x^3+\frac{g^2{x}^2}{r^2}-\frac{2gx}{r^2}+\frac{1}{r^2}=0$ 

$\therefore r^2{x}^3+q^2{x}^2-2qx+1=0$ 
নির্ণেয় সমীকরণ, $r^2{x}^3+q^2{x}^2-2qx+1=0$.