Question

\[5 x^2 - 6x + 2 = 0\]

Answer 1

Given:

\[5 x^2 - 6x + 2 = 0\]

  Comparing the given equation with general form of the quadratic equation 

\[a x^2 + bx + c = 0\], we get 

\[a = 5, b = - 6\] and \[c = 2\].

Substituting these values in

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

\[\alpha = \frac{6 + \sqrt{36 - 4 \times 5 \times 2}}{2 \times 5}\] and \[\beta = \frac{6 - \sqrt{36 - 4 \times 2 \times 5}}{2 \times 5}\]

\[\Rightarrow \alpha = \frac{6 + \sqrt{- 4}}{10}\]     and   \[\beta = \frac{6 - \sqrt{- 4}}{10}\]

\[\Rightarrow \alpha = \frac{6 + \sqrt{4 i^2}}{10}\] and \[\beta = \frac{6 - \sqrt{4 i^2}}{10}\]

\[\Rightarrow \alpha = \frac{6 + 2i}{10}\]  and     \[\beta = \frac{6 - 2i}{10}\]

\[\Rightarrow \alpha = \frac{2 ( 3 + i)}{10}\] and \[\beta = \frac{2 ( 3 - i)}{10}\]

\[\Rightarrow \alpha = \frac{3}{5} + \frac{1}{5}i\]  and    \[\beta = \frac{3}{5} - \frac{1}{5}i\]

Hence, the roots of the equation are \[\frac{3}{5} \pm \frac{1}{5}i .\]