Solve sqrt{u^2+21}-2=u+1

$$\sqrt { u ^ { 2 } + 21 } - 2 = u + 1$$

Solve for u

$$u=2$$

Solution Steps

Subtract -2 from both sides of the equation.

$$\sqrt{u^{2}+21}=u+1+2$$

Add 1 and 2 to get 3.

$$\sqrt{u^{2}+21}=u+3$$

Square both sides of the equation.

$$\left(\sqrt{u^{2}+21}\right)^{2}=\left(u+3\right)^{2}$$

Calculate \sqrt{u^{2}+21} to the power of 2 and get u^{2}+21.

$$u^{2}+21=\left(u+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(u+3\right)^{2}.

$$u^{2}+21=u^{2}+6u+9$$

Subtract u^{2} from both sides.

$$u^{2}+21-u^{2}=6u+9$$

Combine u^{2} and -u^{2} to get 0.

$$21=6u+9$$

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

$$6u+9=21$$

Subtract 9 from both sides.

$$6u=21-9$$

Subtract 9 from 21 to get 12.

$$6u=12$$

Divide both sides by 6.

$$u=\frac{12}{6}$$

Divide 12 by 6 to get 2.

$$u=2$$

Substitute 2 for u in the equation \sqrt{u^{2}+21}-2=u+1.

$$\sqrt{2^{2}+21}-2=2+1$$

Simplify. The value u=2 satisfies the equation.

$$3=3$$

Equation \sqrt{u^{2}+21}=u+3 has a unique solution.

$$u=2$$