সৃজনশীলঃ ‘Px2+qx+1=0……… (i)x3−11×2+47x−85=0…(ii)…’ ময়মনসিংহ বোর্ড ২০২১ – বহুপদী ও বহুপদী সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

প্রদত্ত সমীকরণ: ${\mathrm{Px}}^2+\mathrm{qx}+1=0......\left(i\right)$

ধরি, সমীকরণের মূলদ্বয় $\alpha ,\beta$

$\therefore \alpha +\beta =-\frac{q}{P}$ এবং $\alpha \beta =\frac{1}{p}$

$\Rightarrow P\alpha +P\beta =-q$

$\Rightarrow P\alpha +q=-P\beta$

আবার, $P\alpha +P\beta =-q$

$\Rightarrow P\beta +q=-p\alpha$

$\therefore L.H.S=(P\alpha +q{)}^{-3}+(P\beta +q{)}^{-3}$

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

$=\frac{1}{(-P\beta {)}^3}+\frac{1}{(-P\alpha {)}^3}=\frac{1}{-P^3{\beta }^3}+\frac{1}{-P^3{\alpha }^3}$

$=-\frac{1}{P^3}(\frac{1}{{\beta }^3}+\frac{1}{{\alpha }^3})=-\frac{1}{P^3}\left(\frac{{\alpha }^3+P^3}{{\alpha }^3{P}^3}\right)$

$=-\frac{1}{P^3}\left\{\frac{(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )}{(\alpha \beta {)}^3}\right\}$

$=-\frac{1}{p^3}\left\{\frac{\left(\frac{-q_3}{p}\right)-3\frac{1}{p}(-\frac{q}{p})}{{\left(\frac{1}{p}\right)}^3}\right\}$

$=-\frac{1}{p^3}\times \frac{\frac{-q^3}{p^3}+\frac{3q}{p^2}}{\frac{1}{p^3}}=-\frac{1}{p^3}\times \frac{\frac{-q^3+3\mathrm{pq}}{p^3}}{\frac{1}{p^3}}$

$=-\frac{1}{p^3}\{-q(q^2-3p)\}$

$=\frac{q(q^2-3p)}{p^3}=$ R.H.S (দেখানো হলো)

প্রদত্ত (ii) নং সমীকরণ:

$x^3-11x^2+47x-85=0$

$\therefore 5+\alpha +\beta =11\Rightarrow \alpha +\beta =11-5=6$

$\therefore$ 5. $\alpha \cdot \beta =85\Rightarrow \alpha \beta =17$

$\therefore$ wb‡Y©q mgxKiY:

$x^2-($ মূলদ্বয়ের যোগফল $)x+$ মূলদ্বয়ের গুণফল $=0$

$\Rightarrow x^2-(\alpha +\frac{1}{\beta }+\beta +\frac{1}{\alpha })x+(\alpha +\frac{1}{\beta })(\beta +\frac{1}{\alpha })=0$

$\Rightarrow x^2-\{\frac{\alpha +\beta }{\alpha \beta }+(\alpha +\beta \left)\right\}x+(\alpha \beta +1+1+\frac{1}{\alpha \beta })=0$

$\Rightarrow x^2-(\frac{6}{17}+6)x+(17+2+\frac{1}{17})=0$

$\Rightarrow x^2-\frac{108}{17}x+\frac{324}{17}=0$

$\therefore 17x^2-108x+324=0$