দ্বিঘাত সমীকরণের একটি মূল $\frac{1}{1+\sqrt{-3}}$ হলে সমীকরণটি হবে-

$\begin{aligned} & \frac{1}{1+\sqrt{-3}} \\ = & \frac{1}{1+\sqrt{3} i} \\ = & \frac{1(1-\sqrt{3} i)}{(1+\sqrt{3} i)(1-\sqrt{3} i)} \\ = & \frac{1-\sqrt{3} i}{1-\sqrt{3} i+\sqrt{3} i-3 i^{2}} \\ = & \frac{1-\sqrt{3} i}{1-3 i^{2}} \\ = & \frac{1-\sqrt{3} i}{1+3} \\ = & \frac{1-\sqrt{3} i}{4} \\ = & \frac{1}{4}-\frac{\sqrt{3}}{4} i \end{aligned}$

অন্যমূলটি $\frac{1}{4}+\frac{\sqrt{3}}{4} i$

আমরা জানি,

$x^{2}-$ (মূল এর যোগফল) + মূলের গুনফল = 0

$\begin{array}{l}=x^{2}-\left(\frac{1}{4}+\frac{\sqrt{3}}{4} i+\frac{1}{4}-\frac{\sqrt{3}}{4} i\right) x+\left(\frac{1}{4}+\frac{\sqrt{3}}{4}\right)\left(\frac{1}{4}-\frac{\sqrt{3}}{4} i\right) \\ =x^{2}-\left(\frac{1}{4}+\frac{1}{4}\right) x+\frac{1}{16}-\frac{3}{16} i^{2}=0 \\ =x^{2}-\left(\frac{1+1}{4}\right) x+\frac{1}{16}+\frac{3}{16}=0 \\ =x^{2}-\frac{2}{4} x+\frac{1+3}{16}=0 \\ =x^{2}-\frac{1}{2} x+\frac{4}{16}=0 \\\end{array}$

$\begin{array}{l}x^{2}-\frac{1}{2} x+\frac{1}{4}=0 \\ 4 x^{2}-\frac{4}{2} x+1=0 \\ 4 x^{2}-2 x+1=0\end{array}$