y^2+11left|xright|+3sqrt{x^2+9}=y+2x କୁ ସମାଧାନ କରନ୍ତୁ

$${ y }^{ 2 } +11 \left| x \right| +3 \sqrt{ { x }^{ 2 } +9 } =y+2x$$

x ପାଇଁ ସମାଧାନ କରନ୍ତୁ

$$\left\{\begin{matrix}x=\frac{\sqrt{y^{4}-2y^{3}+y^{2}+648}}{24}-\frac{y^{2}}{8}+\frac{y}{8}\text{, }&\frac{\sqrt{y^{4}-2y^{3}+y^{2}+648}}{24}-\frac{y^{2}}{8}+\frac{y}{8}\geq 0\text{ and }\frac{\sqrt{y^{4}-2y^{3}+y^{2}+648}}{24}-\frac{y^{2}}{72}+\frac{y}{72}\leq 0\text{ and }y^{4}-2y^{3}+y^{2}+648\geq 0\\x=-\frac{\sqrt{y^{4}-2y^{3}+y^{2}+648}}{24}-\frac{y^{2}}{8}+\frac{y}{8}\text{, }&-\frac{\sqrt{y^{4}-2y^{3}+y^{2}+648}}{24}-\frac{y^{2}}{8}+\frac{y}{8}\geq 0\text{ and }-\frac{\sqrt{y^{4}-2y^{3}+y^{2}+648}}{24}-\frac{y^{2}}{72}+\frac{y}{72}\leq 0\text{ and }y^{4}-2y^{3}+y^{2}+648\geq 0\\x=\frac{3\sqrt{y^{4}-2y^{3}+y^{2}+1440}+13y^{2}-13y}{160}\text{, }&\frac{3\sqrt{y^{4}-2y^{3}+y^{2}+1440}+13y^{2}-13y}{160}\leq 0\text{ and }\frac{3\sqrt{y^{4}-2y^{3}+y^{2}+1440}}{160}+\frac{9y^{2}}{2080}-\frac{9y}{2080}\geq 0\text{ and }y^{4}-2y^{3}+y^{2}+1440\geq 0\\x=\frac{-3\sqrt{y^{4}-2y^{3}+y^{2}+1440}+13y^{2}-13y}{160}\text{, }&\frac{-3\sqrt{y^{4}-2y^{3}+y^{2}+1440}+13y^{2}-13y}{160}\leq 0\text{ and }-\frac{3\sqrt{y^{4}-2y^{3}+y^{2}+1440}}{160}+\frac{9y^{2}}{2080}-\frac{9y}{2080}\geq 0\text{ and }y^{4}-2y^{3}+y^{2}+1440\geq 0\end{matrix}\right.$$