Question:
If u2 + (u−2v−1)2 = −4v(u + v), then what is the value of u + 3v?
- 1/4
- 0
- 1/2
- -1/4
Correct Answer: 4
$u^{2}+(u-2 v-1)^{2}=-4 v(u+v)$
$\Rightarrow u^{2}+u^{2}+4 v^{2}+1-4 u v+4 v-2 u+4 v u+4 v^{2}=0$
$\Rightarrow 2 u^{2}-2 u+8 v^{2}+4 v+1=0$
$\Rightarrow 2\left(u^{2}-u+\frac{1}{4}\right)+2\left(4 v^{2}+2 v+\frac{1}{4}\right)=0$
$\Rightarrow 2\left(u-\frac{1}{2}\right)^{2}+2\left(2 v+\frac{1}{2}\right)^{2}=0$
$\Rightarrow u-\frac{1}{2}=0 ; 2 v+\frac{1}{2}=0$
$\mathrm{u}=\frac{1}{2}$ and $\mathrm{v}=-\frac{1}{4}$
$\mathrm{u}+3 \mathrm{v}=\frac{1}{2}-\frac{3}{4}=-\frac{1}{4}$
$u^{2}+(u-2 v-1)^{2}=-4 v(u+v)$
$\Rightarrow u^{2}+u^{2}+4 v^{2}+1-4 u v+4 v-2 u+4 v u+4 v^{2}=0$
$\Rightarrow 2 u^{2}-2 u+8 v^{2}+4 v+1=0$
$\Rightarrow 2\left(u^{2}-u+\frac{1}{4}\right)+2\left(4 v^{2}+2 v+\frac{1}{4}\right)=0$
$\Rightarrow 2\left(u-\frac{1}{2}\right)^{2}+2\left(2 v+\frac{1}{2}\right)^{2}=0$
$\Rightarrow u-\frac{1}{2}=0 ; 2 v+\frac{1}{2}=0$
$\mathrm{u}=\frac{1}{2}$ and $\mathrm{v}=-\frac{1}{4}$
$\mathrm{u}+3 \mathrm{v}=\frac{1}{2}-\frac{3}{4}=-\frac{1}{4}$
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