Question

Найдите координаты пересечения графиков функции: y=3x2 и y=x + 2

Answer 1

$y = 3 {x}^{2} \\ y = x + 2 \\ \\ 3 {x}^{2} = x + 2 \\ 3 {x}^{2} - x - 2 = 0 \\ D = ( - 1) {}^{2} - 4 \times 3 \times ( - 2) = 1 + 24 = 25 \: \: (5) \\ x _{1} = \frac{1 + 5}{6} = 1 \\ x _{2} = \frac{1 - 5}{6} = - \frac{4}{6} = - \frac{2}{3} \\ \\ y _{1} = 1 + 2 = 3 \\ y _{2} = - \frac{2}{3} + 2 = 1 \frac{1}{3} \\ \\ Ответ: \: (1 \: ; \: 3) \: ; \: ( - \frac{2}{3} \: ; \: 1\frac{1}{3} )$