Solve sqrt{x^2+1}=x+1

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

Solve for x

$$x=0$$

Solution Steps

Square both sides of the equation.

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

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

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

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

$$x^{2}+1=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$x^{2}+1-x^{2}=2x+1$$

Combine x^{2} and -x^{2} to get 0.

$$1=2x+1$$

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

$$2x+1=1$$

Subtract 1 from both sides.

$$2x=1-1$$

Subtract 1 from 1 to get 0.

$$2x=0$$

Product of two numbers is equal to 0 if at least one of them is 0. Since 2 is not equal to 0, x must be equal to 0.

$$x=0$$

Substitute 0 for x in the equation \sqrt{x^{2}+1}=x+1.

$$\sqrt{0^{2}+1}=0+1$$

Simplify. The value x=0 satisfies the equation.

$$1=1$$

Equation \sqrt{x^{2}+1}=x+1 has a unique solution.

$$x=0$$