Q^2*y=x^2+3x+4 шешу

$$Q ^ { 2 } \cdot y = x ^ { 2 } + 3 x + 4$$

Q мәнін табыңыз (complex solution)

$$\left\{\begin{matrix}Q=-iy^{-\frac{1}{2}}\sqrt{-x^{2}-3x-4}\text{; }Q=iy^{-\frac{1}{2}}\sqrt{-x^{2}-3x-4}\text{, }&y\neq 0\\Q\in \mathrm{C}\text{, }&\left(x=\frac{-3+\sqrt{7}i}{2}\text{ or }x=\frac{-\sqrt{7}i-3}{2}\right)\text{ and }y=0\end{matrix}\right.$$

x мәнін табыңыз (complex solution)

$$x=\frac{\sqrt{4yQ^{2}-7}-3}{2}$$

$$x=\frac{-\sqrt{4yQ^{2}-7}-3}{2}$$

Q мәнін табыңыз

$$Q=\sqrt{\frac{x^{2}+3x+4}{y}}$$

$$Q=-\sqrt{\frac{x^{2}+3x+4}{y}}\text{, }y>0$$

x мәнін табыңыз

$$x=\frac{\sqrt{4yQ^{2}-7}-3}{2}$$

$$x=\frac{-\sqrt{4yQ^{2}-7}-3}{2}\text{, }y\geq \frac{7}{4Q^{2}}\text{ or }Q=0$$