Question

\[13 x^2 + 7x + 1 = 0\]

Answer 1

Given: 

\[13 x^2 + 7x + 1 = 0\]

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

\[a x^2 + bx + c = 0\], we get 

\[a = 13, b = 7\]  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:

\[\alpha = \frac{- 7 + \sqrt{49 - 4 \times 13 \times 1}}{2 \times 13}\] and \[\beta = \frac{- 7 - \sqrt{49 - 4 \times 13 \times 1}}{2 \times 13}\]

\[\Rightarrow \alpha = \frac{- 7 + \sqrt{49 - 52}}{26}\]   and   \[\beta = \frac{- 7 - \sqrt{49 - 52}}{26}\]

\[\Rightarrow \alpha = \frac{- 7 + \sqrt{- 3}}{26}\] and \[\beta = \frac{- 7 - \sqrt{- 3}}{26}\]

\[\Rightarrow \alpha = \frac{- 7 + i\sqrt{3}}{26}\] and \[\beta = \frac{- 7 - i\sqrt{3}}{26}\]

\[\Rightarrow \alpha = - \frac{7}{26} + \frac{\sqrt{3}}{26}i\] and   \[\beta = - \frac{7}{26} - \frac{\sqrt{3}}{26}i\]

Hence, the roots of the equation are  

\[- \frac{7}{26} \pm \frac{\sqrt{3}}{26}i .\]