$u² + (u − 2v − 1)² = −4v (u + v)$. Then the value of $u + 3v$ is

Given equation: $u^2 + (u - 2v - 1)^2 = -4v(u + v)$

---

We expand $(u - 2v - 1)^2$:

$(u - 2v - 1)^2 = u^2 - 4uv - 2u + 4v^2 + 4v + 1$

We use the formula $(a - b - c)^2 = a^2 + b^2 + c^2 - 2ab - 2ac + 2bc$

So our equation becomes:

$u^2 + u^2 - 4uv - 2u + 4v^2 + 4v + 1 = -4v(u + v)$

---

We expand the right side:

$-4v(u + v) = -4vu - 4v^2$

---

We move everything to one side:

$u^2 + u^2 - 4uv - 2u + 4v^2 + 4v + 1 + 4vu + 4v^2 = 0$

Combining like terms:

$u^2$ terms: $u^2 + u^2 = 2u^2$

$v^2$ terms: $4v^2 + 4v^2 = 8v^2$

$uv$ terms: $-4uv + 4vu = 0$ (they cancel out!)

$u$ terms: $-2u$

$v$ terms: $4v$

Constant: $1$

This gives us: $2u^2 - 2u + 8v^2 + 4v + 1 = 0$

---

We factor out 2 from the first two terms:

$2(u^2 - u) + 8v^2 + 4v + 1 = 0$

We want to complete the square for both $u$ and $v$ terms. Let's rewrite this strategically:

$2(u^2 - u + \tfrac{1}{4}) + 2(4v^2 + 2v + \tfrac{1}{4}) = 0$

To complete the square:

For $u^2 - u$: we add $(\tfrac{1}{2})^2 = \tfrac{1}{4}$

For $4v^2 + 2v$: we need $(2v + \tfrac{1}{2})^2 = 4v^2 + 2v + \tfrac{1}{4}$

This gives us: $2(u - \tfrac{1}{2})^2 + 2(2v + \tfrac{1}{2})^2 = 0$

---

Since we have a sum of squares equal to zero:

$2(u - \tfrac{1}{2})^2 + 2(2v + \tfrac{1}{2})^2 = 0$

A sum of squares can only equal zero if each square equals zero individually (since squares are always non-negative).

Therefore:

$(u - \tfrac{1}{2})^2 = 0 \rightarrow u = \tfrac{1}{2}$

$(2v + \tfrac{1}{2})^2 = 0 \rightarrow 2v + \tfrac{1}{2} = 0 \rightarrow v = -\tfrac{1}{4}$

---

We calculate $u + 3v$:

$u + 3v = \tfrac{1}{2} + 3(-\tfrac{1}{4}) = \tfrac{1}{2} - \tfrac{3}{4} = \tfrac{2}{4} - \tfrac{3}{4} = -\tfrac{1}{4}$

---

Answer: $u + 3v = -\tfrac{1}{4}$

When you see an equation that can be rearranged into a sum of squares equal to zero, remember that each square must individually equal zero. This is a powerful technique for solving systems of equations!