Solve (x-3)^2=ax+bx+9

$$( x - 3 ) ^ { 2 } = a x + b x + 9$$

Solve for a (complex solution)

$$\left\{\begin{matrix}\\a=x-b-6\text{, }&\text{unconditionally}\\a\in \mathrm{C}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9=ax+bx+9$$

Swap sides so that all variable terms are on the left hand side.

$$ax+bx+9=x^{2}-6x+9$$

Subtract bx from both sides.

$$ax+9=x^{2}-6x+9-bx$$

Subtract 9 from both sides.

$$ax=x^{2}-6x+9-bx-9$$

Subtract 9 from 9 to get 0.

$$ax=x^{2}-6x-bx$$

The equation is in standard form.

$$xa=x^{2}-bx-6x$$

Divide both sides by x.

$$\frac{xa}{x}=\frac{x\left(x-b-6\right)}{x}$$

Dividing by x undoes the multiplication by x.

$$a=\frac{x\left(x-b-6\right)}{x}$$

Divide x\left(-6+x-b\right) by x.

$$a=x-b-6$$

Solve for b (complex solution)

$$\left\{\begin{matrix}\\b=x-a-6\text{, }&\text{unconditionally}\\b\in \mathrm{C}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9=ax+bx+9$$

Swap sides so that all variable terms are on the left hand side.

$$ax+bx+9=x^{2}-6x+9$$

Subtract ax from both sides.

$$bx+9=x^{2}-6x+9-ax$$

Subtract 9 from both sides.

$$bx=x^{2}-6x+9-ax-9$$

Subtract 9 from 9 to get 0.

$$bx=x^{2}-6x-ax$$

The equation is in standard form.

$$xb=x^{2}-ax-6x$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{x\left(x-a-6\right)}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{x\left(x-a-6\right)}{x}$$

Divide x\left(-6+x-a\right) by x.

$$b=x-a-6$$

Solve for a

$$\left\{\begin{matrix}\\a=x-b-6\text{, }&\text{unconditionally}\\a\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9=ax+bx+9$$

Swap sides so that all variable terms are on the left hand side.

$$ax+bx+9=x^{2}-6x+9$$

Subtract bx from both sides.

$$ax+9=x^{2}-6x+9-bx$$

Subtract 9 from both sides.

$$ax=x^{2}-6x+9-bx-9$$

Subtract 9 from 9 to get 0.

$$ax=x^{2}-6x-bx$$

The equation is in standard form.

$$xa=x^{2}-bx-6x$$

Divide both sides by x.

$$\frac{xa}{x}=\frac{x\left(x-b-6\right)}{x}$$

Dividing by x undoes the multiplication by x.

$$a=\frac{x\left(x-b-6\right)}{x}$$

Divide x\left(-6+x-b\right) by x.

$$a=x-b-6$$

Solve for b

$$\left\{\begin{matrix}\\b=x-a-6\text{, }&\text{unconditionally}\\b\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9=ax+bx+9$$

Swap sides so that all variable terms are on the left hand side.

$$ax+bx+9=x^{2}-6x+9$$

Subtract ax from both sides.

$$bx+9=x^{2}-6x+9-ax$$

Subtract 9 from both sides.

$$bx=x^{2}-6x+9-ax-9$$

Subtract 9 from 9 to get 0.

$$bx=x^{2}-6x-ax$$

The equation is in standard form.

$$xb=x^{2}-ax-6x$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{x\left(x-a-6\right)}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{x\left(x-a-6\right)}{x}$$

Divide x\left(-6+x-a\right) by x.

$$b=x-a-6$$