Solve x+2=sqrt{4-x^2}

$$x+2= \sqrt{ 4- { x }^{ 2 } }$$

Solve for x

$$x=-2$$

$$x=0$$

Solution Steps

Square both sides of the equation.

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

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

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

Calculate \sqrt{4-x^{2}} to the power of 2 and get 4-x^{2}.

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Add x^{2} to both sides.

$$x^{2}+4x+x^{2}=0$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+4x=0$$

Factor out x.

$$x\left(2x+4\right)=0$$

To find equation solutions, solve x=0 and 2x+4=0.

$$x=0 x=-2$$

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

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

Simplify. The value x=0 satisfies the equation.

$$2=2$$

Substitute -2 for x in the equation x+2=\sqrt{4-x^{2}}.

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

Simplify. The value x=-2 satisfies the equation.

$$0=0$$

List all solutions of x+2=\sqrt{4-x^{2}}.

$$x=0 x=-2$$