Solve sqrt{x}+sqrt{y}=sqrt{a}

$$\sqrt { x } + \sqrt { y } = \sqrt { a }$$

Solve for a

$$a=\left(\sqrt{x}+\sqrt{y}\right)^{2}$$

$$x\geq 0\text{ and }y\geq 0$$

Steps for Solving Radical Equations

Swap sides so that all variable terms are on the left hand side.

$$\sqrt{a}=\sqrt{x}+\sqrt{y}$$

Square both sides of the equation.

$$a=\left(\sqrt{x}+\sqrt{y}\right)^{2}$$

Solve for x

$$x=\left(\sqrt{y}-\sqrt{a}\right)^{2}$$

$$y\geq 0\text{ and }a\geq 0\text{ and }-\left(\sqrt{y}-\sqrt{a}\right)\geq 0$$

Steps for Solving Radical Equations

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

$$\sqrt{x}+\sqrt{y}-\sqrt{y}=\sqrt{a}-\sqrt{y}$$

Subtracting \sqrt{y} from itself leaves 0.

$$\sqrt{x}=\sqrt{a}-\sqrt{y}$$

Square both sides of the equation.

$$x=\left(\sqrt{a}-\sqrt{y}\right)^{2}$$

Solve for a (complex solution)

$$a=\left(\sqrt{x}+\sqrt{y}\right)^{2}$$

$$\left(x=0\text{ and }y=0\right)\text{ or }arg(\sqrt{x}+\sqrt{y})<\pi$$

Solve for x (complex solution)

$$x=\left(\sqrt{y}-\sqrt{a}\right)^{2}$$

$$y=a\text{ or }arg(\sqrt{y}-\sqrt{a})\geq \pi$$