Question

Решите систему уравнений: (5х2 - y2 + 6x = 11; х2 + y2 = 25.​

Answer 1

$\begin{cases}5x^2-y^2+6x=11\\x^2+y^2=25\end{cases}\Longleftrightarrow\quad\begin{cases}y^2=25-x^2\\5x^2-(25-x^2)+6x=11\end{cases}$

$5x^2-(25-x^2)+6x=11\\5x^2-25+x^2+6x=11\\6x^2+6x-25-11=0\\6x^2+6x-36=0\quad|:6$

$x^2+x-6=0\\\\D=1^2-4(-6)=1+24=25\qquad\qquad\boxed{D = b^2-4ac}$

25 > 0  ⇒  2 корня

$\sqrt{25}=5\\\\\left\begin{array}{lcl}x_1=\dfrac{-1-5}{2}=-3\\\\x_2=\dfrac{-1+5}{2}=2\end{array}\right \qquad\qquad\boxed{x=\dfrac{-b\underline+\sqrt{D}}{2a}}$

$y^2=25-2^2=21\\\\y_1=-\sqrt{21}\\y_2=\sqrt{21}$

$y^2=25-(-3)^2=25-9=16\\\\y_3=-\sqrt{16}=-4\\y_4=\sqrt{16}=4$

Ответ: $(-3;\ -4);\ (-3;\ 4);\ (2;\ -\sqrt{21});\ (2;\ \sqrt{21})$