{l}{alpha+b^2-c=0}{b-4a=c}{c-b+a=3} सोडवा

$$\left. \begin{array} { l } { \alpha + b ^ { 2 } - c = 0 } \\ { b - 4 a = c } \\ { c - b + a = 3 } \end{array} \right.$$

α, b, c, a साठी सोडवा (जटिल उत्तर)

$$\alpha \in \mathrm{C}\text{, }b=c-4\text{, }c=\frac{\sqrt{17-4\alpha }+9}{2}\text{, }a=-1$$

$$\alpha \in \mathrm{C}\text{, }b=c-4\text{, }c=\frac{-\sqrt{17-4\alpha }+9}{2}\text{, }a=-1$$

α, b, c, a साठी सोडवा

$$\alpha \leq \frac{17}{4}\text{, }b=c-4\text{, }c=\frac{\sqrt{17-4\alpha }+9}{2}\text{, }a=-1$$

$$\alpha \leq \frac{17}{4}\text{, }b=c-4\text{, }c=\frac{-\sqrt{17-4\alpha }+9}{2}\text{, }a=-1$$