Solve (x-4)^2+b^2=x^2-8

$$( x - 4 ) ^ { 2 } + b ^ { 2 } = x ^ { 2 } - 8$$

Solve for x

$$x=\frac{b^{2}+24}{8}$$

Steps for Solving Linear Equation

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

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

Subtract x^{2} from both sides.

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

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

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

Subtract 16 from both sides.

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

Subtract 16 from -8 to get -24.

$$-8x+b^{2}=-24$$

Subtract b^{2} from both sides.

$$-8x=-24-b^{2}$$

The equation is in standard form.

$$-8x=-b^{2}-24$$

Divide both sides by -8.

$$\frac{-8x}{-8}=\frac{-b^{2}-24}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x=\frac{-b^{2}-24}{-8}$$

Divide -24-b^{2} by -8.

$$x=\frac{b^{2}}{8}+3$$

Solve for b (complex solution)

$$b=-2\sqrt{2\left(x-3\right)}$$

$$b=2\sqrt{2\left(x-3\right)}$$

Solve for b

$$b=2\sqrt{2\left(x-3\right)}$$

$$b=-2\sqrt{2\left(x-3\right)}\text{, }x\geq 3$$