Solve quady=2sqrt{x-1}

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

Solve for x

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

$$y\geq 0$$

Steps for Solving Radical Equations

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

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Square both sides of the equation.

$$x-1=\frac{y^{2}}{4}$$

Add 1 to both sides of the equation.

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

Subtracting -1 from itself leaves 0.

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

Subtract -1 from \frac{y^{2}}{4}.

$$x=\frac{y^{2}}{4}+1$$

Solve for x (complex solution)

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

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

Solve for y

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

$$x\geq 1$$