Solve sqrt{4x+1}=x+1

$$\sqrt { 4 x + 1 } = x + 1$$

Solve for x

$$x=2$$

$$x=0$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{4x+1}\right)^{2}=\left(x+1\right)^{2}$$

Calculate \sqrt{4x+1} to the power of 2 and get 4x+1.

$$4x+1=\left(x+1\right)^{2}$$

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

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

Subtract x^{2} from both sides.

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

Subtract 2x from both sides.

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

Combine 4x and -2x to get 2x.

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

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

Factor out x.

$$x\left(2-x\right)=0$$

To find equation solutions, solve x=0 and 2-x=0.

$$x=0 x=2$$

Substitute 0 for x in the equation \sqrt{4x+1}=x+1.

$$\sqrt{4\times 0+1}=0+1$$

Simplify. The value x=0 satisfies the equation.

$$1=1$$

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

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

Simplify. The value x=2 satisfies the equation.

$$3=3$$

List all solutions of \sqrt{4x+1}=x+1.

$$x=0 x=2$$