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

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

Solve for x

$$x=4$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract x^{2} from both sides.

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

Add 2x to both sides.

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

Combine 2x and 2x to get 4x.

$$4x+1-x^{2}=1$$

Subtract 1 from both sides.

$$4x+1-x^{2}-1=0$$

Subtract 1 from 1 to get 0.

$$4x-x^{2}=0$$

Factor out x.

$$x\left(4-x\right)=0$$

To find equation solutions, solve x=0 and 4-x=0.

$$x=0 x=4$$

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

$$\sqrt{2\times 0+1}=0-1$$

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

$$1=-1$$

Substitute 4 for x in the equation \sqrt{2x+1}=x-1.

$$\sqrt{2\times 4+1}=4-1$$

Simplify. The value x=4 satisfies the equation.

$$3=3$$

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

$$x=4$$