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

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

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

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

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

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

Calculate 4 to the power of 2 and get 16.

$$x^{2}+4x+4=\left(\sqrt{x^{2}+16}\right)^{2}$$

Calculate \sqrt{x^{2}+16} to the power of 2 and get x^{2}+16.

$$x^{2}+4x+4=x^{2}+16$$

Subtract x^{2} from both sides.

$$x^{2}+4x+4-x^{2}=16$$

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

$$4x+4=16$$

Subtract 4 from both sides.

$$4x=16-4$$

Subtract 4 from 16 to get 12.

$$4x=12$$

Divide both sides by 4.

$$x=\frac{12}{4}$$

Divide 12 by 4 to get 3.

$$x=3$$

Substitute 3 for x in the equation x+2=\sqrt{x^{2}+4^{2}}.

$$3+2=\sqrt{3^{2}+4^{2}}$$

Simplify. The value x=3 satisfies the equation.

$$5=5$$

Equation x+2=\sqrt{x^{2}+16} has a unique solution.

$$x=3$$