If $$U^{2}+(U-2V-1)^{2}$$= −$$4V(U+V)$$ , then what is the value of $$U+3V$$ ?

Given that $$U^{2}+(U-2V-1)^{2}$$= −$$4V(U+V)$$ 

$$\Rightarrow$$ $$U^{2}+(U-2V-1)(U-2V-1)$$= −$$4V(U+V)$$

$$\Rightarrow$$ $$U^{2}+(U^2-2UV-U-2UV+4V^2+2V-U+2V+1)$$ = −$$4V(U+V)$$ 

$$\Rightarrow$$ $$U^{2}+(U^2-4UV-2U+4V^2+4V+1)$$ = −$$4V(U+V)$$ 

$$\Rightarrow$$ $$2U^2-4UV-2U+4V^2+4V+1=−4UV-4V^2$$

$$\Rightarrow$$ $$2U^2-2U+8V^2+4V+1=0$$

$$\Rightarrow$$ $$2[U^2-U+\dfrac{1}{4}]+8[V^2+\dfrac{V}{2}+\dfrac{1}{16}]=0$$

$$\Rightarrow$$ $$2(U-\dfrac{1}{2})^2+8(V+\dfrac{1}{4})^2=0$$

Sum of two square terms is zero i.e. individual square term is equal to zero. 

$$U-\dfrac{1}{2}$$ = 0 and $$V+\dfrac{1}{4}$$ = 0

U = $$\dfrac{1}{2}$$ and V = $$-\dfrac{1}{4}$$

Therefore, $$U+3V$$ = $$\dfrac{1}{2}$$+$$\dfrac{-1*3}{4}$$ = $$\dfrac{-1}{4}$$. Hence, option C is the correct answer.