Solve a_{n}=10-2sqrt{2n}+1

$$a _ { n } = 10 - 2 \sqrt { 2 n } + 1$$

Solve for n

$$n=\frac{\left(11-a_{n}\right)^{2}}{8}$$

$$a_{n}\leq 11$$

Steps for Solving Radical Equations

Swap sides so that all variable terms are on the left hand side.

$$10-2\sqrt{2n}+1=a_{n}$$

Subtract 1 from both sides.

$$10-2\sqrt{2n}=a_{n}-1$$

Subtract 10 from both sides.

$$-2\sqrt{2n}=a_{n}-1-10$$

Subtract 10 from -1 to get -11.

$$-2\sqrt{2n}=a_{n}-11$$

Divide both sides by -2.

$$\frac{-2\sqrt{2n}}{-2}=\frac{a_{n}-11}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$\sqrt{2n}=\frac{a_{n}-11}{-2}$$

Divide a_{n}-11 by -2.

$$\sqrt{2n}=\frac{11-a_{n}}{2}$$

Square both sides of the equation.

$$2n=\frac{\left(11-a_{n}\right)^{2}}{4}$$

Divide both sides by 2.

$$\frac{2n}{2}=\frac{\left(11-a_{n}\right)^{2}}{2\times 4}$$

Dividing by 2 undoes the multiplication by 2.

$$n=\frac{\left(11-a_{n}\right)^{2}}{2\times 4}$$

Divide \frac{\left(-a_{n}+11\right)^{2}}{4} by 2.

$$n=\frac{\left(11-a_{n}\right)^{2}}{8}$$

Solve for a_n

$$a_{n}=-2\sqrt{2n}+11$$

$$n\geq 0$$