Question

\[21 x^2 + 9x + 1 = 0\]

Answer 1

Given:  

\[21 x^2 + 9x + 1 = 0\]

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

\[a x^2 + bx + c = 0\] we get 

\[a = 21, b = 9\] and \[c = 1\] .

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

\[\alpha = \frac{- 9 + \sqrt{81 - 4 \times 21 \times 1}}{2 \times 21}\]  and   \[\beta = \frac{- 9 - \sqrt{81 - 4 \times 21 \times 1}}{2 \times 21}\]

\[\Rightarrow \alpha = \frac{- 9 + \sqrt{3}i}{42}\] and \[\beta = \frac{- 9 - \sqrt{3}i}{42}\]

\[\Rightarrow \alpha = - \frac{9}{42} + \frac{\sqrt{3}i}{42}\]  and     \[\beta = - \frac{9}{42} - \frac{\sqrt{3}i}{42}\]

\[\Rightarrow \alpha = - \frac{3}{14} + \frac{\sqrt{3}i}{42}\] and \[\beta = - \frac{3}{14} - \frac{\sqrt{3}i}{42}\] 

Hence, the roots of the equation are \[- \frac{3}{14} \pm \frac{i\sqrt{3}}{42} .\]