Solve x² + (x-2)² = 2(x+1)² : Multiple Squared Terms Equation

Solve the following equation. First, we'll simplify the algebraic expressions using the square of binomial formula:

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

Apply the mentioned formula and expand the parentheses in the expressions in the equation. On the right side, since we have parentheses with an exponent multiplier, we'll expand the (existing) parentheses using the square of binomial formula into additional parentheses (marked with an underline in the following calculation):

$x^2+(x-2)^2=2\underline{(x+1)^2} \\ x^2+x^2-2\cdot x\cdot2+2^2=2\underline{(x^2+2\cdot x\cdot1+1^2)} \\ x^2+x^2-4x+4=2(x^2+2x+1) \\ x^2+x^2-4x+4=2x^2+4x+2 \\$In the final stage, we expand the parentheses on the right side by using the distributive property,

Continue to combine like terms, by moving terms between sides. We observe that the squared term cancels out, therefore it's a first-degree equation, which we'll solve by isolating the variable term on one side and dividing both sides of the equation by its coefficient:

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

In the final stage, we simplified the fraction that was obtained as the solution for $x$.

Therefore, the correct answer is answer B.