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If $u^2+(u-2v-1)^2=-4v(u+v)$, then what is the value of $u+3v$ ?

1. $1/4$
2. $0$
3. $1/2$
4. $-1/4$

Given that, $u^{2} + (u-2v-1)^{2} = -4v (u+v) \quad \longrightarrow (1)$

We know that , $(a-b-c)^{2} = a^{2}+b^{2}+c^{2}-2ab+2bc-2ca$

Now, $u^{2}+u^{2}+4v^{2}+1-4uv+4v-2u=-4uv-4v^{2}$

$\Rightarrow 2u^{2}+8v^{2}+4v-2u+1=0$

$\Rightarrow 2u^{2}-2u+8v^{2}+4v+1=0$

$\Rightarrow 2(u^{2} – u) + 2(4v^{2} + 2v) + 1 = 0$

$\Rightarrow (u^{2} - u) + (4v^{2} + 2v) = – \frac{1}{2}$

$\Rightarrow u^{2} – u + \frac{1}{4} – \frac{1}{4} + (4v^{2} + 2v + \frac{1}{4} – \frac{1}{4}) = – \frac{1}{2}$

$\Rightarrow \left (u – \frac{1}{2} \right)^{2} – \frac{1}{4} + \left(2v + \frac{1}{2} \right)^{2} – \frac{1}{4} = – \frac {1}{2} \quad [ \because (a-b)^{2} = a^{2} + b^{2} – 2ab ,(a+b)^{2} = a^{2} + b^{2} + 2ab ]$

$\Rightarrow \left(u – \frac{1}{2} \right)^{2} + \left(2v + \frac {1}{2} \right)^{2} = – \frac{1}{2} + \frac{1}{2}$

$\Rightarrow \left(u – \frac{1}{2} \right)^{2} + \left(2v + \frac {1}{2} \right)^{2} = 0$

$\Rightarrow \left(u – \frac{1}{2} \right)^{2} =0, \left(2v + \frac {1}{2} \right)^{2} = 0$

$\Rightarrow \boxed {u = \frac{1}{2}}, 2v + \frac{1}{2} = 0 \Rightarrow \boxed{v = \frac{-1}{4}}$

$\therefore$ The value of $u+3v = \frac{1}{2} + 3 \left( \frac{-1}{4} \right) = \frac{1}{2} – \frac{3}{4} = \frac{2-3}{4} = \frac {-1}{4}$

Correct Answer $: \text {D}$
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