Solve a(x-1)^2+b(x+1)^2+c=x^2

$$a ( x - 1 ) ^ { 2 } + b ( x + 1 ) ^ { 2 } + c = x ^ { 2 }$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=-\frac{bx^{2}-x^{2}+2bx+b+c}{\left(x-1\right)^{2}}\text{, }&x\neq 1\\a\in \mathrm{C}\text{, }&b=\frac{1-c}{4}\text{ and }x=1\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$a\left(x^{2}-2x+1\right)+b\left(x+1\right)^{2}+c=x^{2}$$

Use the distributive property to multiply a by x^{2}-2x+1.

$$ax^{2}-2ax+a+b\left(x+1\right)^{2}+c=x^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$ax^{2}-2ax+a+b\left(x^{2}+2x+1\right)+c=x^{2}$$

Use the distributive property to multiply b by x^{2}+2x+1.

$$ax^{2}-2ax+a+bx^{2}+2bx+b+c=x^{2}$$

Subtract bx^{2} from both sides.

$$ax^{2}-2ax+a+2bx+b+c=x^{2}-bx^{2}$$

Subtract 2bx from both sides.

$$ax^{2}-2ax+a+b+c=x^{2}-bx^{2}-2bx$$

Subtract b from both sides.

$$ax^{2}-2ax+a+c=x^{2}-bx^{2}-2bx-b$$

Subtract c from both sides.

$$ax^{2}-2ax+a=x^{2}-bx^{2}-2bx-b-c$$

Reorder the terms.

$$ax^{2}-2ax+a=-bx^{2}+x^{2}-2bx-b-c$$

Combine all terms containing a.

$$\left(x^{2}-2x+1\right)a=-bx^{2}+x^{2}-2bx-b-c$$

Divide both sides by x^{2}-2x+1.

$$\frac{\left(x^{2}-2x+1\right)a}{x^{2}-2x+1}=\frac{-bx^{2}+x^{2}-2bx-b-c}{x^{2}-2x+1}$$

Dividing by x^{2}-2x+1 undoes the multiplication by x^{2}-2x+1.

$$a=\frac{-bx^{2}+x^{2}-2bx-b-c}{x^{2}-2x+1}$$

Divide -bx^{2}+x^{2}-2bx-b-c by x^{2}-2x+1.

$$a=\frac{-bx^{2}+x^{2}-2bx-b-c}{\left(x-1\right)^{2}}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=-\frac{ax^{2}-x^{2}-2ax+a+c}{\left(x+1\right)^{2}}\text{, }&x\neq -1\\b\in \mathrm{C}\text{, }&a=\frac{1-c}{4}\text{ and }x=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$a\left(x^{2}-2x+1\right)+b\left(x+1\right)^{2}+c=x^{2}$$

Use the distributive property to multiply a by x^{2}-2x+1.

$$ax^{2}-2ax+a+b\left(x+1\right)^{2}+c=x^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$ax^{2}-2ax+a+b\left(x^{2}+2x+1\right)+c=x^{2}$$

Use the distributive property to multiply b by x^{2}+2x+1.

$$ax^{2}-2ax+a+bx^{2}+2bx+b+c=x^{2}$$

Subtract ax^{2} from both sides.

$$-2ax+a+bx^{2}+2bx+b+c=x^{2}-ax^{2}$$

Add 2ax to both sides.

$$a+bx^{2}+2bx+b+c=x^{2}-ax^{2}+2ax$$

Subtract a from both sides.

$$bx^{2}+2bx+b+c=x^{2}-ax^{2}+2ax-a$$

Subtract c from both sides.

$$bx^{2}+2bx+b=x^{2}-ax^{2}+2ax-a-c$$

Reorder the terms.

$$bx^{2}+2bx+b=-ax^{2}+x^{2}+2ax-a-c$$

Combine all terms containing b.

$$\left(x^{2}+2x+1\right)b=-ax^{2}+x^{2}+2ax-a-c$$

Divide both sides by x^{2}+2x+1.

$$\frac{\left(x^{2}+2x+1\right)b}{x^{2}+2x+1}=\frac{-ax^{2}+x^{2}+2ax-a-c}{x^{2}+2x+1}$$

Dividing by x^{2}+2x+1 undoes the multiplication by x^{2}+2x+1.

$$b=\frac{-ax^{2}+x^{2}+2ax-a-c}{x^{2}+2x+1}$$

Divide -ax^{2}+x^{2}+2ax-a-c by x^{2}+2x+1.

$$b=\frac{-ax^{2}+x^{2}+2ax-a-c}{\left(x+1\right)^{2}}$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{bx^{2}-x^{2}+2bx+b+c}{\left(x-1\right)^{2}}\text{, }&x\neq 1\\a\in \mathrm{R}\text{, }&b=\frac{1-c}{4}\text{ and }x=1\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$a\left(x^{2}-2x+1\right)+b\left(x+1\right)^{2}+c=x^{2}$$

Use the distributive property to multiply a by x^{2}-2x+1.

$$ax^{2}-2ax+a+b\left(x+1\right)^{2}+c=x^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$ax^{2}-2ax+a+b\left(x^{2}+2x+1\right)+c=x^{2}$$

Use the distributive property to multiply b by x^{2}+2x+1.

$$ax^{2}-2ax+a+bx^{2}+2bx+b+c=x^{2}$$

Subtract bx^{2} from both sides.

$$ax^{2}-2ax+a+2bx+b+c=x^{2}-bx^{2}$$

Subtract 2bx from both sides.

$$ax^{2}-2ax+a+b+c=x^{2}-bx^{2}-2bx$$

Subtract b from both sides.

$$ax^{2}-2ax+a+c=x^{2}-bx^{2}-2bx-b$$

Subtract c from both sides.

$$ax^{2}-2ax+a=x^{2}-bx^{2}-2bx-b-c$$

Reorder the terms.

$$ax^{2}-2ax+a=-bx^{2}+x^{2}-2bx-b-c$$

Combine all terms containing a.

$$\left(x^{2}-2x+1\right)a=-bx^{2}+x^{2}-2bx-b-c$$

Divide both sides by x^{2}-2x+1.

$$\frac{\left(x^{2}-2x+1\right)a}{x^{2}-2x+1}=\frac{-bx^{2}+x^{2}-2bx-b-c}{x^{2}-2x+1}$$

Dividing by x^{2}-2x+1 undoes the multiplication by x^{2}-2x+1.

$$a=\frac{-bx^{2}+x^{2}-2bx-b-c}{x^{2}-2x+1}$$

Divide -bx^{2}+x^{2}-2bx-b-c by x^{2}-2x+1.

$$a=\frac{-bx^{2}+x^{2}-2bx-b-c}{\left(x-1\right)^{2}}$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{ax^{2}-x^{2}-2ax+a+c}{\left(x+1\right)^{2}}\text{, }&x\neq -1\\b\in \mathrm{R}\text{, }&a=\frac{1-c}{4}\text{ and }x=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$a\left(x^{2}-2x+1\right)+b\left(x+1\right)^{2}+c=x^{2}$$

Use the distributive property to multiply a by x^{2}-2x+1.

$$ax^{2}-2ax+a+b\left(x+1\right)^{2}+c=x^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$ax^{2}-2ax+a+b\left(x^{2}+2x+1\right)+c=x^{2}$$

Use the distributive property to multiply b by x^{2}+2x+1.

$$ax^{2}-2ax+a+bx^{2}+2bx+b+c=x^{2}$$

Subtract ax^{2} from both sides.

$$-2ax+a+bx^{2}+2bx+b+c=x^{2}-ax^{2}$$

Add 2ax to both sides.

$$a+bx^{2}+2bx+b+c=x^{2}-ax^{2}+2ax$$

Subtract a from both sides.

$$bx^{2}+2bx+b+c=x^{2}-ax^{2}+2ax-a$$

Subtract c from both sides.

$$bx^{2}+2bx+b=x^{2}-ax^{2}+2ax-a-c$$

Reorder the terms.

$$bx^{2}+2bx+b=-ax^{2}+x^{2}+2ax-a-c$$

Combine all terms containing b.

$$\left(x^{2}+2x+1\right)b=-ax^{2}+x^{2}+2ax-a-c$$

Divide both sides by x^{2}+2x+1.

$$\frac{\left(x^{2}+2x+1\right)b}{x^{2}+2x+1}=\frac{-ax^{2}+x^{2}+2ax-a-c}{x^{2}+2x+1}$$

Dividing by x^{2}+2x+1 undoes the multiplication by x^{2}+2x+1.

$$b=\frac{-ax^{2}+x^{2}+2ax-a-c}{x^{2}+2x+1}$$

Divide -ax^{2}+x^{2}+2ax-a-c by x^{2}+2x+1.

$$b=\frac{-ax^{2}+x^{2}+2ax-a-c}{\left(x+1\right)^{2}}$$