সৃজনশীলঃ ‘∫(x)=ax2+bx+c, g(x)=p2+qx+r…’ ময়মনসিংহ বোর্ড ২০২১ – বহুপদী ও বহুপদী সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $\int \left(x\right)=0$

$\Rightarrow {\mathrm{ax}}^2+\mathrm{bx}+c=0......\left(i\right)$

সমীকরণের মূলদ্বয় $\alpha ,\beta$ হলে, $\alpha +\beta =-\frac{b}{a}$ ও $\alpha \beta =\frac{c}{a}$

আবার, $a^2{x}^2-(b^2-2ac)x+c^2=0..........\left(ii\right)$

$\Rightarrow x^2-(\frac{b^2}{a^2}-2\frac{c}{a})x+\frac{c^2}{a^2}=0[a^2$ দ্বারা ভাগ করে]

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

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

$\Rightarrow x^2-{\alpha }^{2}x-{\beta }^{2}x+{\alpha }^2{\beta }^2=0$

$begin{aligned}& Rightarrow mathrm{x}left(mathrm{x}-alpha^{2}right)-beta^{2}left(mathrm{x}-alpha^{2}right)=0 & Rightarrowleft(mathrm{x}-alpha^{2}right)left(mathrm{x}-beta^{2}right)=0 & therefore mathrm{x}=alpha^{2} quad text { অথবা } quad mathrm{x}=beta^{2}end{aligned}$

$\therefore$ সমীকরণের মূলদ্বয় ${\alpha }^2$ ও ${\beta }^2$

দেওয়া আছে, $f\left(x\right)=0;a=2,c=3,p=5,r=7$

$\Rightarrow {\mathrm{ax}}^2+\mathrm{bx}+c=0$

$\Rightarrow 2x^2+\mathrm{bx}+3=0.........\left(i\right)$

ধরি, (i) নং সমীকরণের মূলদ্বয় $\alpha ,\beta$

$\therefore \alpha +\beta =-\frac{b}{2}\alpha \beta =\frac{3}{2}$

আবার, $g\left(x\right)=0$

$\Rightarrow {\mathrm{px}}^2+\mathrm{qx}+r=0$

$\Rightarrow 5x+\mathrm{qx}+7=0........\left(ii\right)$

ধরি, (ii) নং সমীকরণের মূলদ্বয় $\gamma ,\delta$

$\therefore \gamma +\delta =-\frac{q}{5}$ এবং $\gamma \delta =\frac{7}{5}$

প্রশ্নমতে, $\frac{\alpha }{\beta }=\frac{\gamma }{\delta }\Rightarrow \frac{\alpha +\beta }{\alpha -\beta }=\frac{\gamma +\delta }{\gamma -\delta }$

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

$\Rightarrow \frac{(\alpha +\beta {)}^2}{(\alpha -\beta {)}^2-4\alpha \beta }=\frac{(\gamma +\delta {)}^2}{(\gamma +\delta {)}^2-4\gamma \delta }$

$\Rightarrow \frac{(\alpha +\beta {)}^2}{(\gamma +\delta {)}^2}=\frac{(\alpha +\beta {)}^2-4\alpha \beta }{(\gamma +\delta {)}^2-4\gamma \delta }$

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

$\begin{array}{cc}& \Rightarrow \frac{{\left(\frac{-b}{2}\right)}^2}{{\left(\frac{-q}{5}\right)}^2}=\frac{\frac{3}{2}}{\frac{7}{5}}\Rightarrow \frac{\frac{b^2}{4}}{\frac{q^2}{25}}=\frac{15}{14}\\ & \Rightarrow \frac{b^2}{q^2}=\frac{15}{14}\times \frac{4}{25}=\frac{6}{35}\end{array}$

$\therefore b:q=\sqrt{6}:\sqrt{35}$