If $$ x $$ and $$ y $$ are real numbers such that
$$
x^2 + (x - 2y - 1)^2 = -4y(x + y),
$$
then the value of $$ x - 2y $$ is:

Question

If $$ x $$ and $$ y $$ are real numbers such that  
$$
x^2 + (x - 2y - 1)^2 = -4y(x + y),
$$  
then the value of $$ x - 2y $$ is:

Accepted Answer

To solve the given equation:

**Expand and simplify the equation:**  
   The given equation is:  
   $$
   x^2 + (x - 2y - 1)^2 = -4y(x + y).
   $$  
   Expanding $$ (x - 2y - 1)^2 $$:  
   $$
   x^2 + (x^2 - 4xy - 4x + 4y^2 + 4y + 1) = -4yx - 4y^2.
   $$  
   Combine like terms:  
   $$
   2x^2 + (-4xy - 4x) + (4y^2 + 4y + 1) = -4yx - 4y^2.
   $$

**Factorize and simplify further:**  
   Rearrange terms to form:  
   $$
   (x + 2y)^2 + (x - 2y - 1)^2 = 0.
   $$

**Analyze the equation:**  
   Since a square of a real number is always non-negative, both $$ (x + 2y)^2 = 0 $$ and $$ (x - 2y - 1)^2 = 0 $$.  
   This implies:  
   $$
   x + 2y = 0, \quad x - 2y - 1 = 0.
   $$

**Solve for $$ x $$ and $$ y $$:**  
   From $$ x + 2y = 0 $$, we get $$ x = -2y $$.  
   Substituting into $$ x - 2y = 1 $$:  
   $$
   (-2y) - 2y = -1.
   $$

Thus, the value of  x - 2y = **1**.

Answer

Answer

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