সৃজনশীলঃ ‘P(x)=ax2+bx+c(a±0) এবং Q(x)=mx2+nx+n(m±0)…’ ২০২৫ – জটিল সংখ্যা, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $p\left(x\right)=a{x}^2+bx+c$ 
এখানে, $p\left(x\right)=0$ বा, ${\mathrm{ax}}^2+\mathrm{bx}+c=0$ 
$a{x}^2+bx+c=0$ সমীকরণের মূলদ্বয় $\gamma ,\delta$ 
$\therefore$ মূলদ্বয়ের যোগফল, $\gamma +\delta =-\frac{b}{a}$ 
বা, $a\gamma +b=-a\delta$ এবং $a\delta +b=-ay$ 
মূলদ্বয়ের গুণফল, $\gamma \beta =\frac{c}{a}$, 

বামপক্ষ $=(a\gamma +b{)}^{-2}+(a\delta +b{)}^{-2}$ 

$\begin{array}{cc}& =\frac{1}{(a\gamma +b{)}^2}+\frac{1}{(a\delta +b{)}^2}\\ & =\frac{1}{(-a\delta {)}^2}+\frac{1}{(-a\gamma {)}^2}=\frac{1}{a^2{\delta }^2}+\frac{1}{a^2{\gamma }^2}=\frac{{\gamma }^2+{\delta }^2}{a^2{\gamma }^2{\delta }^2}\\ & =\frac{(\gamma +\delta {)}^2-2\gamma \delta }{a^2(\gamma \delta {)}^2}=\frac{{(-\frac{b}{a})}^2-2\frac{c}{a}}{a^2{\left(\frac{c}{a}\right)}^2}=\frac{\frac{b^2}{a^2}-2\frac{c}{a}}{c^2}\end{array}$ 

$=\frac{b^2-2\mathrm{ac}}{a^2{c}^2}=$ ডানপক্ষ 

$\therefore (a\gamma +b{)}^{-2}+(a\delta +b{)}^{-2}=\frac{b^2-2ac}{a^2{c}^2}$ (প্রমাণিত) 

দেওয়া আছে, $Q\left(x\right)={\mathrm{mx}}^2+\mathrm{nx}+n$$Q\left(x\right)=0$ হলে, $m{x}^2+nx+n=0$ সমীকরণটির  মূলদ্বয়ের অনুপাত  p:2q মনে করি, মূলদ্বয়  $p\alpha$ ও $2q\alpha$ 

$\therefore p\alpha +2q\alpha =-\frac{n}{m}$ 

বा, $\alpha (p+2q)=\frac{-n}{m}$ 

$\therefore p+2q=\frac{-n}{m\alpha }$ 

এবং $p\alpha \times 2q\alpha =\frac{n}{m}$
বा, $2pq{\alpha }^2=\frac{n}{m}$ 

$\therefore \mathrm{pq}=\frac{n}{2m{\alpha }^2}$ 

বামপক্ষ $=\sqrt{\frac{p}{q}}+2\sqrt{\frac{q}{p}}+\sqrt{\frac{2n}{m}}$ 

$\begin{array}{cc}& =\frac{\sqrt{p}}{\sqrt{q}}+\frac{2\sqrt{q}}{\sqrt{p}}+\sqrt{\frac{2n}{m}}=\frac{(\sqrt{p}{)}^2+2(\sqrt{q}{)}^2}{\sqrt{pq}}+\sqrt{\frac{2n}{m}}\\ & =\frac{p+2q}{\sqrt{pq}}+\sqrt{\frac{2n}{m}}=\frac{\frac{-n}{m\alpha }}{\frac{\sqrt{n}}{\sqrt{2m{\alpha }^2}}}+\sqrt{\frac{2n}{m}}\\ & =\frac{-n}{m\alpha }\times \frac{\sqrt{2m}\alpha }{\sqrt{n}}+\sqrt{\frac{2n}{m}}\\ & =\frac{-\sqrt{n}\cdot \sqrt{n}}{\sqrt{m}\sqrt{m}\cdot \alpha }\times \frac{\sqrt{2}\sqrt{m}}{\sqrt{n}}+\sqrt{\frac{2n}{m}}\\ & =\frac{-\sqrt{2}\times \sqrt{n}}{\sqrt{m}}+\sqrt{\frac{2n}{m}}=-\sqrt{\frac{2n}{m}}+\sqrt{\frac{2n}{m}}\end{array}$ 

$=0=$ ডানপক্ষ 

$\therefore \sqrt{\frac{p}{q}}+2\sqrt{\frac{q}{p}}+\sqrt{\frac{2n}{m}}=0$. (দেখানো হলো)