Question

\[\sqrt{3} x^2 - \sqrt{2}x + 3\sqrt{3} = 0\]

Answer 1

Given:

\[\sqrt{3} x^2 - \sqrt{2}x + 3\sqrt{3} = 0\]

Comparing the given equation with the general form of the quadratic equation 

\[a x^2 + bx + c = 0\], we get 

\[a = \sqrt{3}, b = - \sqrt{2}\]  and   \[c = 3\sqrt{3}\].

Substituting these values in 

\[\alpha = \frac{- b + \sqrt{b^2 - 4ac}}{2a}\] and \[\beta = \frac{- b - \sqrt{b^2 - 4ac}}{2a}\], we get:

\[\alpha = \frac{\sqrt{2} + \sqrt{2 - 4 \times \sqrt{3} \times 3\sqrt{3}}}{2\sqrt{3}}\] and   \[\beta = \frac{\sqrt{2} - \sqrt{2 - 4 \times \sqrt{3} \times 3\sqrt{3}}}{2\sqrt{3}}\]

\[\Rightarrow \alpha = \frac{\sqrt{2} + \sqrt{- 34}}{2\sqrt{3}}\] and \[\beta = \frac{\sqrt{2} - \sqrt{- 34}}{2\sqrt{3}}\]

\[\Rightarrow \alpha = \frac{\sqrt{2} + i\sqrt{34}}{2\sqrt{3}}\]  and  \[\beta = \frac{\sqrt{2} - i\sqrt{34}}{2\sqrt{3}}\]  

Hence, the roots of the equation are 

\[\frac{\sqrt{2} \pm i\sqrt{34}}{2\sqrt{3}}\].