Solve x(x+a)+36=(x+b)^2

$$x ( x + a ) + 36 = ( x + b ) ^ { 2 }$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{2bx+b^{2}-36}{x}\text{, }&x\neq 0\\a\in \mathrm{C}\text{, }&\left(b=-6\text{ or }b=6\right)\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x by x+a.

$$x^{2}+xa+36=\left(x+b\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(x+b\right)^{2}.

$$x^{2}+xa+36=x^{2}+2xb+b^{2}$$

Subtract x^{2} from both sides.

$$xa+36=x^{2}+2xb+b^{2}-x^{2}$$

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

$$xa+36=2xb+b^{2}$$

Subtract 36 from both sides.

$$xa=2xb+b^{2}-36$$

The equation is in standard form.

$$xa=2bx+b^{2}-36$$

Divide both sides by x.

$$\frac{xa}{x}=\frac{2bx+b^{2}-36}{x}$$

Dividing by x undoes the multiplication by x.

$$a=\frac{2bx+b^{2}-36}{x}$$

Solve for a

$$\left\{\begin{matrix}a=\frac{2bx+b^{2}-36}{x}\text{, }&x\neq 0\\a\in \mathrm{R}\text{, }&x=0\text{ and }|b|=6\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x by x+a.

$$x^{2}+xa+36=\left(x+b\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(x+b\right)^{2}.

$$x^{2}+xa+36=x^{2}+2xb+b^{2}$$

Subtract x^{2} from both sides.

$$xa+36=x^{2}+2xb+b^{2}-x^{2}$$

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

$$xa+36=2xb+b^{2}$$

Subtract 36 from both sides.

$$xa=2xb+b^{2}-36$$

The equation is in standard form.

$$xa=2bx+b^{2}-36$$

Divide both sides by x.

$$\frac{xa}{x}=\frac{2bx+b^{2}-36}{x}$$

Dividing by x undoes the multiplication by x.

$$a=\frac{2bx+b^{2}-36}{x}$$

Solve for b (complex solution)

$$b=-\left(\sqrt{x^{2}+ax+36}+x\right)$$

$$b=\sqrt{x^{2}+ax+36}-x$$