Solve (2^2+x)^2=16+x^2

$$( 2 ^ { 2 } + x ) ^ { 2 } = 16 + x ^ { 2 }$$

Solve for x

$$x=0$$

Steps for Solving Linear Equation

Calculate 2 to the power of 2 and get 4.

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

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

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

Subtract x^{2} from both sides.

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

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

$$16+8x=16$$

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.

$$x=0$$