Solve sqrt{x/z}+sqrt{9y}=sqrt{2}

$$\sqrt{ \frac{ x }{ z } } + \sqrt{ 9y } = \sqrt{ 2 }$$

Solve for x

$$x=-z\left(-9y+6\sqrt{2y}-2\right)$$

$$y\geq 0\text{ and }-3\sqrt{y}+\sqrt{2}\geq 0\text{ and }z\neq 0$$

Steps for Solving Radical Equations

Subtract \sqrt{9y} from both sides of the equation.

$$\sqrt{\frac{1}{z}x}+\sqrt{9y}-\sqrt{9y}=\sqrt{2}-\sqrt{9y}$$

Subtracting \sqrt{9y} from itself leaves 0.

$$\sqrt{\frac{1}{z}x}=\sqrt{2}-\sqrt{9y}$$

Subtract \sqrt{9y} from \sqrt{2}.

$$\sqrt{\frac{1}{z}x}=-3\sqrt{y}+\sqrt{2}$$

Square both sides of the equation.

$$\frac{1}{z}x=\left(-3\sqrt{y}+\sqrt{2}\right)^{2}$$

Divide both sides by z^{-1}.

$$\frac{\frac{1}{z}xz}{1}=\frac{\left(-3\sqrt{y}+\sqrt{2}\right)^{2}z}{1}$$

Dividing by z^{-1} undoes the multiplication by z^{-1}.

$$x=\frac{\left(-3\sqrt{y}+\sqrt{2}\right)^{2}z}{1}$$

Divide \left(\sqrt{2}-3\sqrt{y}\right)^{2} by z^{-1}.

$$x=z\left(-3\sqrt{y}+\sqrt{2}\right)^{2}$$

Solve for y

$$y=\frac{\left(-\sqrt{\frac{x}{z}}+\sqrt{2}\right)^{2}}{9}$$

$$\left(x\leq 0\text{ or }z>0\right)\text{ and }\left(z<0\text{ or }x\geq 0\right)\text{ and }z\neq 0\text{ and }-\sqrt{\frac{x}{z}}+\sqrt{2}\geq 0$$

Steps for Solving Radical Equations

Subtract \sqrt{xz^{-1}} from both sides of the equation.

$$\sqrt{9y}+\sqrt{\frac{x}{z}}-\sqrt{\frac{x}{z}}=\sqrt{2}-\sqrt{\frac{x}{z}}$$

Subtracting \sqrt{xz^{-1}} from itself leaves 0.

$$\sqrt{9y}=\sqrt{2}-\sqrt{\frac{x}{z}}$$

Subtract \sqrt{xz^{-1}} from \sqrt{2}.

$$\sqrt{9y}=-\sqrt{\frac{x}{z}}+\sqrt{2}$$

Square both sides of the equation.

$$9y=\left(-\sqrt{\frac{x}{z}}+\sqrt{2}\right)^{2}$$

Divide both sides by 9.

$$\frac{9y}{9}=\frac{\left(-\sqrt{\frac{x}{z}}+\sqrt{2}\right)^{2}}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$y=\frac{\left(-\sqrt{\frac{x}{z}}+\sqrt{2}\right)^{2}}{9}$$

Solve for x (complex solution)

$$x=-z\left(-9y+6\sqrt{2y}-2\right)$$

$$z\neq 0\text{ and }\left(y=\frac{2}{9}\text{ or }arg(-3\sqrt{y}+\sqrt{2})<\pi \right)$$

Solve for y (complex solution)

$$y=\frac{\left(-\sqrt{\frac{x}{z}}+\sqrt{2}\right)^{2}}{9}$$

$$\left(z=\frac{x}{2}\text{ and }x\neq 0\right)\text{ or }\left(z\neq 0\text{ and }arg(-\sqrt{\frac{x}{z}}+\sqrt{2})<\pi \right)$$