Question

$5x^2-19x+17=0$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ 5x^2-19x+17=0 \\ \mathrm{On}\mathrm{comparing\; it\; with}a{x}^2+bx+c=0,\text{we get:} \\ a=5,b=-19\mathrm{and}c=17 \\ \mathrm{Discriminant}D\mathrm{is}\mathrm{given}\mathrm{by}: \\ D=(b^2-4ac) \\ \text{=}{(-19)}^2-4\times 5\times 17 \\ \text{= 361}-340 \\ =21>0 \\ \mathrm{Hence},\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}\mathrm{are}\mathrm{real}. \\ \mathrm{Roots}\alpha \mathrm{and}\beta \mathit{}\mathrm{are}\mathrm{given}\mathrm{by}: \\ \alpha =\frac{-b+\sqrt{D}}{2a}=\frac{-(-19)+\sqrt{21}}{2\times 5}=\frac{-(-19)+\sqrt{21}}{10}=\frac{19+\sqrt{21}}{10} \\ \beta =\frac{-b-\sqrt{D}}{2a}=\frac{-(-19)-\sqrt{21}}{2\times 5}=\frac{-(-19)-\sqrt{21}}{10}=\frac{19-\sqrt{21}}{10} \\ \mathrm{Thus},\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}\mathrm{are}\frac{19+\sqrt{21}}{10}\text{and}\frac{19-\sqrt{21}}{10}\text{.} \end{gathered}$$