Solve y=f(x)=sqrt{2x-4}

$$y = f ( x ) = \sqrt { 2 x - 4 }$$

Solve for x

$$x=\frac{y^{2}+4}{2}$$

$$y\geq 0$$

Steps for Solving Radical Equations

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

$$\sqrt{2x-4}=y$$

Square both sides of the equation.

$$2x-4=y^{2}$$

Add 4 to both sides of the equation.

$$2x-4-\left(-4\right)=y^{2}-\left(-4\right)$$

Subtracting -4 from itself leaves 0.

$$2x=y^{2}-\left(-4\right)$$

Subtract -4 from y^{2}.

$$2x=y^{2}+4$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{y^{2}+4}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{y^{2}+4}{2}$$

Divide y^{2}+4 by 2.

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

Solve for x (complex solution)

$$x=\frac{y^{2}+4}{2}$$

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

Solve for y (complex solution)

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

Solve for y

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

$$x\geq 2$$