Question

x^2+3xy+2y^2=3 2x^2-xy+y^2=8 система уравнений, помогите решить

Answer 1

$\left \{ {{x^2+3xy+2y^2=3\, |\cdot 8} \atop {2x^2-xy+y^2=8\, |\cdot (-3)}} \right.\; \; \left \{ {{8x^2+24xy+16y^2=24} \atop {-6x^2+3xy-3y^2=-24}} \right. \oplus \; \left \{ {{x^2+3xy+2y^2=3} \atop {2x^2+27xy+13y^2=0}} \right.\\\\\\2x^2+27xy+13y^2=0\, |:y^2\ne 0\\\\2\cdot (\frac{x}{y})^2+27\cdot (\frac{x}{y})+13=0\; \; ,\; \; t=\frac{x}{y}\\\\2t^2+27t+13=0\; \; ,\; \; D=625\; ,\; t_1=-\frac{1}{2}\; \; ,\; \; t_2=-13\\\\a)\; \frac{x}{y}=-\frac{1}{2}\; \; \to \; \; y=-2x\; \; ,\; \; x^2+3xy+2y^2=x^2-6x^2+8x^2=3x^2$

$3x^2=3\; \; \to \; \; x^2=1\; ,\; \; x_1=-1\; ,\; x_2=+1\\\\y_1=-2\cdot (-1)=2\; ,\; \; y_2=-2\cdot 1=-2\\\\b)\; \; \frac{x}{y}=-13\; \; \to \; \; x=-13y\; ,\; \; x^2+3xy+2y^2=169y^2-39y^2+2y^2=132y^2\\\\132y^2=3\; ,\; \; y^2=\frac{1}{44} \; \; \to \; \; y_1=-\frac{1}{2\sqrt{11}}\; ,\; y_2=+\frac{1}{2\sqrt{11}}\\\\x_1=\frac{13}{2\sqrt{11}}\; \; ,\; \; x_2=-\frac{13}{2\sqrt{11}}\\\\Otvet:\; \; (-1,2)\; ,\; (1,-2)\; ,\; (\frac{13}{2\sqrt{11}},- \frac{1}{2\sqrt{11}})\; ,\; (-\frac{13}{2\sqrt{11}},\frac{1}{2\sqrt{11}})\; .$

$P.S.\; \; \frac{1}{2\sqrt{11}}=\frac{\sqrt{11}}{22}\; \; ,\; \; \frac{13}{2\sqrt{11}}=\frac{13\sqrt{11}}{22}\; .$