Solve 4^2+(4-x)^2=(4+x)^2

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

Solve for x

$$x=1$$

Steps for Solving Linear Equation

Calculate 4 to the power of 2 and get 16.

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

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

$$16+16-8x+x^{2}=\left(4+x\right)^{2}$$

Add 16 and 16 to get 32.

$$32-8x+x^{2}=\left(4+x\right)^{2}$$

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

$$32-8x+x^{2}=16+8x+x^{2}$$

Subtract 8x from both sides.

$$32-8x+x^{2}-8x=16+x^{2}$$

Combine -8x and -8x to get -16x.

$$32-16x+x^{2}=16+x^{2}$$

Subtract x^{2} from both sides.

$$32-16x+x^{2}-x^{2}=16$$

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

$$32-16x=16$$

Subtract 32 from both sides.

$$-16x=16-32$$

Subtract 32 from 16 to get -16.

$$-16x=-16$$

Divide both sides by -16.

$$x=\frac{-16}{-16}$$

Divide -16 by -16 to get 1.

$$x=1$$