Solve y=2sqrt{x-1}+3

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

Solve for x

$$x=\frac{\left(3-y\right)^{2}+4}{4}$$

$$\frac{y-3}{2}\geq 0$$

Steps for Solving Radical Equations

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

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

Subtract 3 from both sides.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Square both sides of the equation.

$$x-1=\frac{\left(y-3\right)^{2}}{4}$$

Add 1 to both sides of the equation.

$$x-1-\left(-1\right)=\frac{\left(y-3\right)^{2}}{4}-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$x=\frac{\left(y-3\right)^{2}}{4}-\left(-1\right)$$

Subtract -1 from \frac{\left(y-3\right)^{2}}{4}.

$$x=\frac{\left(y-3\right)^{2}}{4}+1$$

Solve for x (complex solution)

$$x=\frac{\left(3-y\right)^{2}+4}{4}$$

$$y=3\text{ or }arg(\frac{3-y}{2})\geq \pi$$

Solve for y

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

$$x\geq 1$$