Solve the Equation: x²+(x-2)²=2(x+1)² with Square Binomials

Let's examine the given equation:

$x^2+(x-2)^2=2(x+1)^2$

First, let's simplify the equation, for this we'll use the perfect square formula for a binomial squared:

$(a\pm b)^2=a^2\pm2ab+b^2$,

We'll start by opening the parentheses in both sides simultaneously using the perfect square formula, then we'll combine like terms,

Note that according to the order of operations (which prioritizes exponents over multiplication), the expression in the right-hand parentheses is first squared and then the resulting expression is multiplied by 2,

Therefore, the expression we get from applying the perfect square formula to the right-hand side will be put in parentheses which we'll multiply by 2 (highlighted with an underline in the following calculation):

$x^2+(x-2)^2=2\underline{(x+1)^2} \\ \downarrow\\ x^2+x^2-2\cdot x\cdot2+2^2=2\underline{(x^2+2\cdot x\cdot1+1^2)}\\ x^2+x^2-4x+4=2(x^2+2x+1)\\$ Let's continue, first we'll open the parentheses on the right side using the distributive property, move terms and combine like terms, in the final step we'll solve the simplified equation we get:

$x^2+x^2-4x+4=2(x^2+2x+1)\\ 2x^2-4x+4=2x^2+4x+2\\ -8x=-2\hspace{6pt}\text{/:(-8)}\\ x=\frac{-2}{-8}\\ \downarrow\\ \boxed{x=\frac{1}{4}}$

where in the final step we reduced the fraction that we got as the solution for the unknown,

Therefore the correct answer is answer B.