Question

\[21 x^2 - 28x + 10 = 0\]

Answer 1

Given:  

\[21 x^2 - 28x + 10 = 0\]

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

\[a x^2 + bx + c = 0\], we get 

\[a = 21, b = - 28\] and \[c = 10\].

Substituting these values in

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

\[\alpha = \frac{28 + \sqrt{784 - 4 \times 21 \times 10}}{2 \times 21}\] and   \[\beta = \frac{28 - \sqrt{784 - 4 \times 21 \times 10}}{2 \times 21}\]

\[\Rightarrow \alpha = \frac{28 + \sqrt{- 56}}{42}\]  and   \[\beta = \frac{28 - \sqrt{- 56}}{42}\]

\[\Rightarrow \alpha = \frac{28 + 2i\sqrt{14}}{42}\] and   \[\beta = \frac{28 - 2i\sqrt{14}}{42}\]

\[\Rightarrow \alpha = \frac{14 + i\sqrt{14}}{21}\]  and \[\beta = \frac{14 - i\sqrt{14}}{21}\]

\[\Rightarrow \alpha = \frac{2}{3} + \frac{\sqrt{14}}{21}i\] and   \[\beta = \frac{2}{3} - \frac{\sqrt{14}}{21}i\]

Hence, the roots of the equation are 

\[\frac{2}{3} \pm \frac{\sqrt{14}}{21}\].