বীজগণিতের সূত্রাবলী

দুইটি ভ্যারিয়েবলের সূত্র

$$\begin{array}{rl}(a+b{)}^2& =a^2+2ab+b^2\\ (a-b{)}^2& =a^2-2ab+b^2\\ a^2-b^2& =(a+b)(a-b)\end{array}$$

তিনটি ভ্যারিয়েবলের সূত্র

$$\begin{array}{rl}(a-b)(a-c)& =a^2-(b+c)a+bc\end{array}$$

দ্বিঘাত সমীকরণের সমাধান

$a{x}^2+bx+c=0$ হলে,

Click to calculation

$$\begin{array}{rl}a{x}^2+bx+c& =0\\ x^2+{\displaystyle \frac{b}{a}}x+{\displaystyle \frac{c}{a}}& =0\\ x^2+{\displaystyle \frac{b}{a}}x& =-{\displaystyle \frac{c}{a}}\\ x^2+{\displaystyle \frac{b}{a}}x+{\left({\displaystyle \frac{b}{2a}}\right)}^2& =-{\displaystyle \frac{c}{a}}+{\left({\displaystyle \frac{b}{2a}}\right)}^2\\ {(x+{\displaystyle \frac{b}{2a}})}^2& ={\displaystyle \frac{b^2-4ac}{4a^2}}\\ x+{\displaystyle \frac{b}{2a}}& =\pm \sqrt{{\displaystyle \frac{b^2-4ac}{4a^2}}}\\ x& =-{\displaystyle \frac{b}{2a}}\pm {\displaystyle \frac{\sqrt{b^2-4ac}}{2a}}\end{array}$$

$x$ এর মান - $$\begin{array}{r}x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}\end{array}$$

ত্রিঘাত সমীকরণের সমাধান

$a{x}^3+b{x}^2+cx+d=0$ হলে $$\begin{array}{rl}x& =\sqrt[3]{({\displaystyle \frac{-b^3}{27a^3}}+{\displaystyle \frac{bc}{6a^2}}-{\displaystyle \frac{d}{2a}})+\sqrt{{({\displaystyle \frac{-b^3}{27a^3}}+{\displaystyle \frac{bc}{6a^2}}-{\displaystyle \frac{d}{2a}})}^2+{({\displaystyle \frac{c}{3a}}-{\displaystyle \frac{b^2}{9a^2}})}^3}}\\ & +\sqrt[3]{({\displaystyle \frac{-b^3}{27a^3}}+{\displaystyle \frac{bc}{6a^2}}-{\displaystyle \frac{d}{2a}})-\sqrt{{({\displaystyle \frac{-b^3}{27a^3}}+{\displaystyle \frac{bc}{6a^2}}-{\displaystyle \frac{d}{2a}})}^2+{({\displaystyle \frac{c}{3a}}-{\displaystyle \frac{b^2}{9a^2}})}^3}}-{\displaystyle \frac{b}{3a}}\end{array}$$ একটু ছোট করে লেখা যায় এভাবে- ধরো $p={\displaystyle \frac{-b}{3a}},q=p^3+{\displaystyle \frac{bc-3ad}{6a^2}},r={\displaystyle \frac{c}{3a}}$। তাহলে উপরের সমীকরণটি দাড়াচ্ছে- $$\begin{array}{r}x=\{q+[q^2+(r-p^2{)}^3{]}^{1/2}{\}}^{1/3}+\{q-[q^2+(r-p^2{)}^3{]}^{1/2}{\}}^{1/3}\end{array}$$

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