Solve sqrt{x+1}=sqrt{x}+1

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

Solve for x

$$x=0$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{x+1} to the power of 2 and get x+1.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{x}+1\right)^{2}.

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

Calculate \sqrt{x} to the power of 2 and get x.

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

Subtract x from both sides.

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

Combine x and -x to get 0.

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

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

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

$$2\sqrt{x}=0$$

Divide both sides by 2. Zero divided by any non-zero number gives zero.

$$\sqrt{x}=0$$

Square both sides of the equation.

$$x=0$$

Substitute 0 for x in the equation \sqrt{x+1}=\sqrt{x}+1.

$$\sqrt{0+1}=\sqrt{0}+1$$

Simplify. The value x=0 satisfies the equation.

$$1=1$$

Equation \sqrt{x+1}=\sqrt{x}+1 has a unique solution.

$$x=0$$