Solve sqrt{(2x-1)}+y=0

$$\sqrt { ( 2 x - 1 ) } + y = 0$$

Solve for x

$$x=\frac{y^{2}+1}{2}$$

$$y\leq 0$$

Steps for Solving Radical Equations

Subtract y from both sides of the equation.

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

Subtracting y from itself leaves 0.

$$\sqrt{2x-1}=-y$$

Square both sides of the equation.

$$2x-1=y^{2}$$

Add 1 to both sides of the equation.

$$2x-1-\left(-1\right)=y^{2}-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$2x=y^{2}-\left(-1\right)$$

Subtract -1 from y^{2}.

$$2x=y^{2}+1$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{y^{2}+1}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{y^{2}+1}{2}$$

Solve for x (complex solution)

$$x=\frac{y^{2}+1}{2}$$

$$arg(y)\geq \pi \text{ or }y=0$$

Solve for y (complex solution)

$$y=-\sqrt{2x-1}$$

Solve for y

$$y=-\sqrt{2x-1}$$

$$x\geq \frac{1}{2}$$