Question

Решите систему уравнений
x = y'2-3y
x+y=8

Answer 1

Ответ:

$\displaystyle (4;4),\ (10;-2)$

Объяснение:

$\displaystyle \begin{cases}x=y^2-3y\\x+y=8\end{cases}\\\\\begin{cases}x=y^2-3y\\y^2-3y+y=8\end{cases}\\\\\begin{cases}x=y^2-3y\\y^2-2y-8=0\end{cases}$

$\displaystyle y^2-2y-8=0\\\\D=(-2)^2-4\cdot 1\cdot (-8)=4+32=36\\\\\sqrt{D}=\sqrt{36}=6\\\\y_1=\frac{2-6}{2\cdot 1}=\frac{-4}{2}=-2\\\\y_2=\frac{2+6}{2\cdot 1}=\frac{8}{2}=4\\\\x_1-2=8\\\\x_1=8+2\\\\x_1=10\\\\x_2+4=8\\\\x_2=8-4\\\\x_2=4$

$\displaystyle (4;4),\ (10;-2)$