Solve sqrt{x+1}=x-1

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

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract x^{2} from both sides.

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

Add 2x to both sides.

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

Combine x and 2x to get 3x.

$$3x+1-x^{2}=1$$

Subtract 1 from both sides.

$$3x+1-x^{2}-1=0$$

Subtract 1 from 1 to get 0.

$$3x-x^{2}=0$$

Factor out x.

$$x\left(3-x\right)=0$$

To find equation solutions, solve x=0 and 3-x=0.

$$x=0 x=3$$

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

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

Simplify. The value x=0 does not satisfy the equation because the left and the right hand side have opposite signs.

$$1=-1$$

Substitute 3 for x in the equation \sqrt{x+1}=x-1.

$$\sqrt{3+1}=3-1$$

Simplify. The value x=3 satisfies the equation.

$$2=2$$

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

$$x=3$$