সৃজনশীলঃ ‘দৃশ্যকল্প-১: 3×2+4x+7=0 সমীকরণের…’ ময়মনসিংহ বোর্ড ২০২৩ – বহুপদী ও বহুপদী সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে,

$3x^2+4x+7=0$ সমীকরণের মূলদ্বয় $\alpha$ ও $\beta$ 

$\therefore \alpha +\beta =-\frac{4}{3}$ এবং $\alpha \beta =\frac{7}{3}$

এখন, ${\alpha }^{-2}$ ও ${\beta }^{-2}$ মূলবিশিষ্ট সমীকরণ নির্ণয় করতে হবে

মূলদ্বয়ের যোগফল

= ${\alpha }^{-2}+{\beta }^{-2}=\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}$

$=\frac{{\alpha }^2+{\beta }^2}{{\alpha }^2{\beta }^2}$

$=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{(\alpha \beta {)}^2}$

= $\frac{{(-\frac{4}{3})}^2-2\times \frac{7}{3}}{{\left(\frac{7}{3}\right)}^2}$

= $\frac{\frac{16}{9}-\frac{14}{3}}{\frac{49}{9}}=\frac{\frac{16-42}{9}}{\frac{49}{9}}=-\frac{26}{9}\times \frac{9}{49}=-\frac{26}{49}$

এবং মূলদ্বয়ের গুণফল

$={\alpha }^{-2}\times {\beta }^{-2}=\frac{1}{{\alpha }^2\cdot {\beta }^2}$

$=\frac{1}{(\alpha \beta {)}^2}=\frac{1}{{\left(\frac{7}{3}\right)}^2}=\frac{1}{\frac{49}{9}}=\frac{9}{49}$

নির্ণেয় সমীকরণ, $x^2-($ মূলদ্বয়ের যোগফল $)x+$ মূলদ্বয়ের গুণফল $=0$ বা, $x^2-(-\frac{26}{49})x+\frac{9}{49}=0$$\therefore 49x^2+26x+9=0$.

দেওয়া আছে,

$f\left(x\right)=x^3-p{x}^2+qx-r$

এবং $f\left(x\right)=0$

বা, $x^3-p{x}^2+qx-r=0$

সমীকরণটির মূলত্রয়, $\alpha ,\beta ,\gamma$

$\therefore \alpha +\beta +\gamma =-\frac{-p}{1}=p$

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

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

$\therefore \Sigma \frac{1}{{\alpha }^3}=\frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}+\frac{1}{{\gamma }^3}$

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

$=(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}-\frac{1}{\alpha \beta }-\frac{1}{\beta \gamma }-\frac{1}{\gamma \alpha })+\frac{3}{\alpha \beta \gamma }$

$[\because a^3+b^3+c^3-3abc=(a+b+c\left)(a^2+b^2+c^2-ab-bc-ca)\right]$

$=\left(\frac{\beta \gamma +\gamma \alpha +\alpha \beta }{\alpha \beta \gamma }\right)\{{(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })}^2-2(\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha })-(\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha })\}+\frac{3}{\alpha \beta \gamma }$

$[\because a^2+b^2+c^2=(a+b+c{)}^2-2(ab+bc+ca)]$

$=\left(\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right)\{{\left(\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right)}^2-3(\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha })\}+\frac{3}{\alpha \beta \gamma }$

$=\left(\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right)\{{\left(\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right)}^2-3\left(\frac{\alpha +\beta +\gamma }{\alpha \beta \gamma }\right)\}+\frac{3}{\alpha \beta \gamma }$

$=\frac{g}{r}\{{\left(\frac{g}{r}\right)}^2-3\left(\frac{p}{r}\right)\}+\frac{3}{r}$

$=\frac{q}{r}(\frac{q^2}{r^2}-\frac{3p}{r})+\frac{3}{r}=\frac{q^3}{r^3}-\frac{3\mathrm{pq}}{r^2}+\frac{3}{r}=\frac{q^3-3\mathrm{pqr}+3r^2}{r^3}$

নির্ণেয় মান: $\frac{q^3-3pqr+3r^2}{r^3}$.