Question

\[17 x^2 - 8x + 1 = 0\]

Answer 1

Given:    

\[17 x^2 - 8x + 1 = 0\]

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

\[a x^2 + bx + c = 0\], we get

\[a = 17, b = - 8\] 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:

\[\alpha = \frac{8 + \sqrt{64 - 4 \times 17 \times 1}}{2 \times 17}\] and   \[\beta = \frac{8 - \sqrt{64 - 4 \times 17 \times 1}}{2 \times 17}\]

\[\Rightarrow \alpha = \frac{8 + \sqrt{64 - 68}}{34}\] and \[\beta = \frac{8 - \sqrt{64 - 68}}{34}\]

\[\Rightarrow \alpha = \frac{8 + \sqrt{- 4}}{34}\] and \[\beta = \frac{8 - \sqrt{- 4}}{34}\]

\[\Rightarrow \alpha = \frac{8 + \sqrt{4 i^2}}{34}\] and \[\beta = \frac{8 - \sqrt{4 i^2}}{34}\]

\[\Rightarrow \alpha = \frac{8 + 2i}{34}\]  and \[\beta = \frac{8 - 2i}{34}\]

\[\Rightarrow \alpha = \frac{4 + i}{17}\]   and \[\beta = \frac{4 - i}{17}\]

\[\Rightarrow \alpha = \frac{4}{17} + \frac{1}{17}i\]  and   \[\beta = \frac{4}{17} - \frac{1}{17}i\]

Hence, the roots of the equation are \[\frac{4}{17} \pm \frac{1}{17}i\].