Solve sqrt{x^2+16}=x+4

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

Solve for x

$$x=0$$

Solution Steps

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$x^{2}+16=x^{2}+8x+16$$

Subtract x^{2} from both sides.

$$x^{2}+16-x^{2}=8x+16$$

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

$$16=8x+16$$

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

$$8x+16=16$$

Subtract 16 from both sides.

$$8x=16-16$$

Subtract 16 from 16 to get 0.

$$8x=0$$

Product of two numbers is equal to 0 if at least one of them is 0. Since 8 is not equal to 0, x must be equal to 0.

$$x=0$$

Substitute 0 for x in the equation \sqrt{x^{2}+16}=x+4.

$$\sqrt{0^{2}+16}=0+4$$

Simplify. The value x=0 satisfies the equation.

$$4=4$$

Equation \sqrt{x^{2}+16}=x+4 has a unique solution.

$$x=0$$