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

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

Solve for x

$$x=0$$

Steps for Solving Linear Equation

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

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

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

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

Subtract x^{2} from both sides.

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

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

$$-4x+4=4x+4$$

Subtract 4x from both sides.

$$-4x+4-4x=4$$

Combine -4x and -4x to get -8x.

$$-8x+4=4$$

Subtract 4 from both sides.

$$-8x=4-4$$

Subtract 4 from 4 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.

$$x=0$$