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

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

Solve for x

$$x=-\frac{z^{2}}{32}+8$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(16-x\right)^{2}.

$$x^{2}+z^{2}=256-32x+x^{2}$$

Add 32x to both sides.

$$x^{2}+z^{2}+32x=256+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+z^{2}+32x-x^{2}=256$$

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

$$z^{2}+32x=256$$

Subtract z^{2} from both sides.

$$32x=256-z^{2}$$

Divide both sides by 32.

$$\frac{32x}{32}=\frac{256-z^{2}}{32}$$

Dividing by 32 undoes the multiplication by 32.

$$x=\frac{256-z^{2}}{32}$$

Divide -z^{2}+256 by 32.

$$x=-\frac{z^{2}}{32}+8$$

Solve for z (complex solution)

$$z=-4\sqrt{16-2x}$$

$$z=4\sqrt{16-2x}$$

Solve for z

$$z=4\sqrt{16-2x}$$

$$z=-4\sqrt{16-2x}\text{, }x\leq 8$$