Solve sqrt{gammax+r}-sqrt{x+1}=1

$$\sqrt { \gamma x + r } - \sqrt { x + 1 } = 1$$

Solve for r

$$r=-x\gamma +\left(\sqrt{x+1}+1\right)^{2}$$

$$x\geq -1$$

Steps for Solving Radical Equations

Subtract -\sqrt{x+1} from both sides of the equation.

$$\sqrt{r+x\gamma }-\sqrt{x+1}-\left(-\sqrt{x+1}\right)=1-\left(-\sqrt{x+1}\right)$$

Subtracting -\sqrt{x+1} from itself leaves 0.

$$\sqrt{r+x\gamma }=1-\left(-\sqrt{x+1}\right)$$

Subtract -\sqrt{x+1} from 1.

$$\sqrt{r+x\gamma }=\sqrt{x+1}+1$$

Square both sides of the equation.

$$r+x\gamma =\left(\sqrt{x+1}+1\right)^{2}$$

Subtract \gamma x from both sides of the equation.

$$r+x\gamma -x\gamma =\left(\sqrt{x+1}+1\right)^{2}-x\gamma$$

Subtracting \gamma x from itself leaves 0.

$$r=\left(\sqrt{x+1}+1\right)^{2}-x\gamma$$

Subtract \gamma x from \left(1+\sqrt{x+1}\right)^{2}.

$$r=-x\gamma +\left(\sqrt{x+1}+1\right)^{2}$$

Solve for r (complex solution)

$$r=-x\gamma +\left(\sqrt{x+1}+1\right)^{2}$$

$$arg(\sqrt{x+1}+1)<\pi$$