Solve sqrt{2n+1}+1=n

$$\sqrt { 2 n + 1 } + 1 = n$$

Solve for n

$$n=4$$

Solution Steps

Subtract 1 from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

Subtract n^{2} from both sides.

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

Add 2n to both sides.

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

Combine 2n and 2n to get 4n.

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

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

Factor out n.

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

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

$$n=0 n=4$$

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

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

Simplify. The value n=0 does not satisfy the equation.

$$2=0$$

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

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

Simplify. The value n=4 satisfies the equation.

$$4=4$$

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

$$n=4$$