Maharashtra BoardEnglish MediumSTD 11 Commerce / ArtsMaths (commerce)Complex Numbers(p-1)3 Marks
Question
Solve the following quadratic equations. $2 x^2-\sqrt{ } 3 x+1=0$
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Answer
Given equation is $2 x^2-\sqrt{3} x+1=0$ Comparing with $a x^2+b x+c=0$, we get $ a=2, b=-\sqrt{3}, c=1 $ Discriminant $=b^2-4 a c$ $ \begin{aligned} & =(-\sqrt{ } 3)^2-4 \times 2 \times 1 \\ & =3-8 \\ & =-5<0 \end{aligned} $ So, the given equation has complex roots. These roots are given by $ \begin{aligned} x & =\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a}} \\ & =\frac{-(-\sqrt{3}) \pm \sqrt{-5}}{2(2)} \\ x & =\frac{\sqrt{3} \pm \sqrt{5} \mathrm{i}}{4} \text { } \end{aligned} $ $\therefore$ the roots of the given equation are $\frac{\sqrt{3}+\sqrt{5} i}{4}$ and $\frac{\sqrt{3}-\sqrt{5} i}{4}$
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