Solve Qquadb^2+x^2=(8-x)^2

$$Q \quad b ^ { 2 } + x ^ { 2 } = ( 8 - x ) ^ { 2 }$$

Solve for Q (complex solution)

$$\left\{\begin{matrix}Q=-\frac{16\left(x-4\right)}{b^{2}}\text{, }&b\neq 0\\Q\in \mathrm{C}\text{, }&x=4\text{ and }b=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-x\right)^{2}.

$$Qb^{2}+x^{2}=64-16x+x^{2}$$

Subtract x^{2} from both sides.

$$Qb^{2}=64-16x+x^{2}-x^{2}$$

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

$$Qb^{2}=64-16x$$

The equation is in standard form.

$$b^{2}Q=64-16x$$

Divide both sides by b^{2}.

$$\frac{b^{2}Q}{b^{2}}=\frac{64-16x}{b^{2}}$$

Dividing by b^{2} undoes the multiplication by b^{2}.

$$Q=\frac{64-16x}{b^{2}}$$

Divide 64-16x by b^{2}.

$$Q=\frac{16\left(4-x\right)}{b^{2}}$$

Solve for Q

$$\left\{\begin{matrix}Q=-\frac{16\left(x-4\right)}{b^{2}}\text{, }&b\neq 0\\Q\in \mathrm{R}\text{, }&x=4\text{ and }b=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-x\right)^{2}.

$$Qb^{2}+x^{2}=64-16x+x^{2}$$

Subtract x^{2} from both sides.

$$Qb^{2}=64-16x+x^{2}-x^{2}$$

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

$$Qb^{2}=64-16x$$

The equation is in standard form.

$$b^{2}Q=64-16x$$

Divide both sides by b^{2}.

$$\frac{b^{2}Q}{b^{2}}=\frac{64-16x}{b^{2}}$$

Dividing by b^{2} undoes the multiplication by b^{2}.

$$Q=\frac{64-16x}{b^{2}}$$

Divide 64-16x by b^{2}.

$$Q=\frac{16\left(4-x\right)}{b^{2}}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=-4iQ^{-\frac{1}{2}}\sqrt{x-4}\text{; }b=4iQ^{-\frac{1}{2}}\sqrt{x-4}\text{, }&Q\neq 0\\b\in \mathrm{C}\text{, }&x=4\text{ and }Q=0\end{matrix}\right.$$

Solve for b

$$\left\{\begin{matrix}b=4\sqrt{-\frac{x-4}{Q}}\text{; }b=-4\sqrt{-\frac{x-4}{Q}}\text{, }&\left(Q>0\text{ and }x\leq 4\right)\text{ or }\left(x\geq 4\text{ and }Q<0\right)\\b\in \mathrm{R}\text{, }&x=4\text{ and }Q=0\end{matrix}\right.$$