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

We need to find the value of $x - 2y$ given that $x^2 + (x - 2y - 1)^2 = -4y(x + y)$ where $x$ and $y$ are real numbers.

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We move all terms to one side:

$x^2 + (x - 2y - 1)^2 + 4y(x + y) = 0$

We want to create a sum of squares, which will help us use a powerful property.

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Expanding $4y(x + y) = 4yx + 4y^2$:

$x^2 + 4yx + 4y^2 + (x - 2y - 1)^2 = 0$

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The first three terms form a perfect square: $x^2 + 4yx + 4y^2$

This follows the pattern $a^2 + 2ab + b^2 = (a + b)^2$ where $a = x$ and $b = 2y$.

So: $x^2 + 4yx + 4y^2 = x^2 + 2 \cdot x \cdot 2y + (2y)^2 = (x + 2y)^2$

Our equation becomes:

$(x + 2y)^2 + (x - 2y - 1)^2 = 0$

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When the sum of squares equals zero, each square must individually equal zero.

Since $x$ and $y$ are real numbers:

$(x + 2y)^2 \geq 0$ (any real number squared is non-negative)

$(x - 2y - 1)^2 \geq 0$ (any real number squared is non-negative)

If two non-negative numbers add up to zero, both must be zero.

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From $(x + 2y)^2 = 0$: we get $x + 2y = 0$

From $(x - 2y - 1)^2 = 0$: we get $x - 2y - 1 = 0$, which means $x - 2y = 1$

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The value of $x - 2y$ is 1.