{l}{x^2+y^2=8}{z=x+y}{a=z}{b=a}{c=b}{text{Solvefor}dtext{where}}{d=c} ਨੂੰ ਹੱਲ ਕਰੋ

$$\left. \begin{array} { l } { x ^ {2} + y ^ {2} = 8 }\\ { z = x + y }\\ { a = z }\\ { b = a }\\ { c = b }\\ { \text{Solve for } d \text{ where} } \\ { d = c } \end{array} \right.$$

x, y, z ਲਈ ਹਲ ਕਰੋ (ਜਟਿਲ ਹੱਲ)

$$\left\{\begin{matrix}x=\frac{\sqrt{-2c\sqrt{16-c^{2}}+16}}{2}\text{, }y=\frac{\sqrt{16-c^{2}}+c}{2}\text{, }z=c\text{, }&d=c\text{ and }a=c\text{ and }b=c\text{ and }arg(\frac{-\sqrt{16-c^{2}}+c}{2})<\pi \\x=\frac{\sqrt{2c\sqrt{16-c^{2}}+16}}{2}\text{, }y=\frac{-\sqrt{16-c^{2}}+c}{2}\text{, }z=c\text{, }&\left(c=-2\sqrt{2}\text{ and }a=-2\sqrt{2}\text{ and }b=-2\sqrt{2}\text{ and }d=-2\sqrt{2}\right)\text{ or }\left(d=c\text{ and }a=c\text{ and }b=c\text{ and }arg(\frac{\sqrt{16-c^{2}}+c}{2})<\pi \right)\\x=-\frac{\sqrt{-2c\sqrt{16-c^{2}}+16}}{2}\text{, }y=\frac{\sqrt{16-c^{2}}+c}{2}\text{, }z=c\text{, }&d=c\text{ and }a=c\text{ and }b=c\text{ and }arg(\frac{\sqrt{16-c^{2}}-c}{2})<\pi \\x=-\frac{\sqrt{2c\sqrt{16-c^{2}}+16}}{2}\text{, }y=\frac{-\sqrt{16-c^{2}}+c}{2}\text{, }z=c\text{, }&\left(c=-2\sqrt{2}\text{ and }a=-2\sqrt{2}\text{ and }b=-2\sqrt{2}\text{ and }d=-2\sqrt{2}\right)\text{ or }\left(d=c\text{ and }a=c\text{ and }b=c\text{ and }arg(\frac{-\sqrt{16-c^{2}}-c}{2})<\pi \right)\\x=0\text{, }y=2\sqrt{2}\approx 2.828427125\text{, }z=2\sqrt{2}\approx 2.828427125\text{, }&c=2\sqrt{2}\text{ and }a=2\sqrt{2}\text{ and }b=2\sqrt{2}\text{ and }d=2\sqrt{2}\end{matrix}\right.$$

x, y, z ਲਈ ਹਲ ਕਰੋ

$$\left\{\begin{matrix}x=-\frac{\sqrt{2\left(-b\sqrt{16-b^{2}}+8\right)}}{2}\text{, }y=\frac{\sqrt{16-b^{2}}+b}{2}\text{, }z=b\text{, }&b=d\text{ and }a=d\text{ and }c=d\text{ and }d\leq 2\sqrt{2}\text{ and }|d|\leq 4\\x=-\frac{\sqrt{2\left(b\sqrt{16-b^{2}}+8\right)}}{2}\text{, }y=\frac{-\sqrt{16-b^{2}}+b}{2}\text{, }z=b\text{, }&b=d\text{ and }a=d\text{ and }c=d\text{ and }d\geq -4\text{ and }d\leq -2\sqrt{2}\\x=\frac{\sqrt{2\left(-b\sqrt{16-b^{2}}+8\right)}}{2}\text{, }y=\frac{\sqrt{16-b^{2}}+b}{2}\text{, }z=b\text{, }&b=d\text{ and }a=d\text{ and }c=d\text{ and }d\geq 2\sqrt{2}\text{ and }d\leq 4\\x=\frac{\sqrt{2\left(b\sqrt{16-b^{2}}+8\right)}}{2}\text{, }y=\frac{-\sqrt{16-b^{2}}+b}{2}\text{, }z=b\text{, }&b=d\text{ and }a=d\text{ and }c=d\text{ and }d\geq -2\sqrt{2}\text{ and }|d|\leq 4\end{matrix}\right.$$