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

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

Solve for z

$$z=0$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract z^{2} from both sides.

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

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

$$4z+4=4$$

Subtract 4 from both sides.

$$4z=4-4$$

Subtract 4 from 4 to get 0.

$$4z=0$$

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

$$z=0$$

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

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

Simplify. The value z=0 satisfies the equation.

$$2=2$$

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

$$z=0$$