Question

Найти точки пересечения параболы y^2+6y+2x-1=0 с прямой 4x+y-3=0

Answer 1

$\begin{cases}y^2+6y+2x-1=0|*2\\4x+y-3=0=\ \textgreater \ y=3-4x\end{cases}\\(3-4x)^2+6(3-4x)+2x-1=0\\16x^2-24x+9+18-24x+2x-1=0\\16x^2-46x+26=0\\x_{1,2}=\frac{23^+_-\sqrt{113}}{16}\\x_1=\frac{23+\sqrt{113}}{16}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_2=\frac{23-\sqrt{113}}{16}\\y_1=3-\frac{23+\sqrt{113}}{4}=\frac{-11+\sqrt{113}}{4}\ \ \ y_2=3-\frac{23-\sqrt{113}}{4}=\frac{-11-\sqrt{113}}{4}\\OTBET:(\frac{23+\sqrt{113}}{16};\frac{-11+\sqrt{113}}{4}});(\frac{23-\sqrt{113}}{16};\frac{-11-\sqrt{113}}{4}})$