Solve sqrt{x-2sqrt{3}}=y+sqrt{3}

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

Solve for x

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

$$y+\sqrt{3}\geq 0$$

Steps for Solving Radical Equations

Square both sides of the equation.

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

Subtract -2\sqrt{3} from both sides of the equation.

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

Subtracting -2\sqrt{3} from itself leaves 0.

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

Subtract -2\sqrt{3} from \left(y+\sqrt{3}\right)^{2}.

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

Solve for x (complex solution)

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

$$y=-\sqrt{3}\text{ or }arg(y+\sqrt{3})<\pi$$

Solve for y (complex solution)

$$y=\sqrt{x-2\sqrt{3}}-\sqrt{3}$$

Solve for y

$$y=\sqrt{x-2\sqrt{3}}-\sqrt{3}$$

$$x\geq 2\sqrt{3}$$