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

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

Solve for x

$$x=4$$

Solution Steps

Subtract x from both sides of the equation.

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

Square both sides of the equation.

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

Expand \left(-\sqrt{2x+1}\right)^{2}.

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

Calculate -1 to the power of 2 and get 1.

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

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

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

Use the distributive property to multiply 1 by 2x+1.

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

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

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

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

Add 2x to both sides.

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

Combine 2x and 2x to get 4x.

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

Subtract x^{2} from both sides.

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

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

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

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

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

Simplify. The value x=4 satisfies the equation.

$$1=1$$

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

$$x=4$$