Solve the Symmetric Squares Equation: (1 - x)^2 = (2x + 1)^2

To solve the equation $(-x+1)^2=(2x+1)^2$, we will follow these steps:

• Step 1: Recognize the equation as an application of the identity $a^2 = b^2$. This implies $a = b$ or $a = -b$.
• Step 2: Apply the identity to our equation.
• Step 3: Solve each of the resulting equations individually to find the possible values of $x$.

Now, let's perform each step in detail:

Step 1: We have the equation $(-x+1)^2 = (2x+1)^2$. According to the identity $a^2 = b^2$, we can set up the following cases:
Case 1: $-x + 1 = 2x + 1$,
Case 2: $-x + 1 = -(2x + 1)$.

Step 2: Solve Case 1:
From $-x + 1 = 2x + 1$, subtract 1 from both sides: $-x = 2x$.
Adding $x$ to both sides gives $0 = 3x$.
Divide by 3: $x = 0$.

Step 3: Solve Case 2:
From $-x + 1 = -(2x + 1)$, distribute the negative sign on the right: $-x + 1 = -2x - 1$.
Add $2x$ to both sides: $x + 1 = -1$.
Subtract 1 from both sides: $x = -2$.

Therefore, the solutions to the equation are $x_1=0$ and $x_2=-2$.

The correct answer is:

$x_1=0,x_2=-2$