Question

$25x^2+30x+7=0$

Answer 1

$$\begin{gathered} \text{Given:} \\ \text{25}{x}^2+30x+7=0 \\ \mathrm{On}\mathrm{comparing\; it\; with}a{x}^2+bx+c=0,\text{we get:} \\ a=25,b=30\mathrm{and}c=7 \\ \mathrm{Discriminant}D\mathrm{is}\mathrm{given}\mathrm{by}: \\ D=(b^2-4ac) \\ {\text{= 30}}^2-4\times 25\times 7 \\ \text{= 900}-700 \\ \text{= 200} \\ \text{= 200 > 0} \\ \text{Hence, the roots of the equation are real.} \\ \mathrm{Roots}\alpha \mathrm{and}\beta \mathrm{are}\mathrm{given}\mathrm{by}: \\ \alpha =\frac{-b+\sqrt{D}}{2a}=\frac{-30+\sqrt{200}}{2\times 25}=\frac{-30+10\sqrt{2}}{50}=\frac{10(-3+\sqrt{2})}{50}=\frac{(-3+\sqrt{2})}{5} \\ \beta =\frac{-b-\sqrt{D}}{2a}=\frac{-30-\sqrt{200}}{2\times 25}=\frac{-30-10\sqrt{2}}{50}=\frac{10(-3-\sqrt{2})}{50}=\frac{(-3-\sqrt{2})}{5} \\ \mathrm{Thus}\text{, the roots of the equation are}\frac{(-3+\sqrt{2})}{5}\text{and}\frac{(-3-\sqrt{2})}{5}\text{.} \end{gathered}$$