Solve sqrt{a^2+4}=a+2

$$\sqrt{ { a }^{ 2 } +4 } = a+2$$

Solve for a

$$a=0$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{a^{2}+4} to the power of 2 and get a^{2}+4.

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

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

$$a^{2}+4=a^{2}+4a+4$$

Subtract a^{2} from both sides.

$$a^{2}+4-a^{2}=4a+4$$

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

$$4=4a+4$$

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

$$4a+4=4$$

Subtract 4 from both sides.

$$4a=4-4$$

Subtract 4 from 4 to get 0.

$$4a=0$$

Product of two numbers is equal to 0 if at least one of them is 0. Since 4 is not equal to 0, a must be equal to 0.

$$a=0$$

Substitute 0 for a in the equation \sqrt{a^{2}+4}=a+2.

$$\sqrt{0^{2}+4}=0+2$$

Simplify. The value a=0 satisfies the equation.

$$2=2$$

Equation \sqrt{a^{2}+4}=a+2 has a unique solution.

$$a=0$$