ແກ້ u^2+4(1+v)^2=1

$$u ^ { 2 } + 4 ( 1 + v ) ^ { 2 } = 1$$

ແກ້ສຳລັບ u (complex solution)

$$u=-\sqrt{-4v^{2}-8v-3}$$

$$u=\sqrt{-4v^{2}-8v-3}$$

ແກ້ສຳລັບ v (complex solution)

$$v=-\frac{\sqrt{1-u^{2}}}{2}-1$$

$$v=\frac{\sqrt{1-u^{2}}}{2}-1$$

ແກ້ສຳລັບ u

$$u=\sqrt{-4v^{2}-8v-3}$$

$$u=-\sqrt{-4v^{2}-8v-3}\text{, }v\geq -\frac{3}{2}\text{ and }v\leq -\frac{1}{2}$$

ແກ້ສຳລັບ v

$$v=-\frac{\sqrt{1-u^{2}}}{2}-1$$

$$v=\frac{\sqrt{1-u^{2}}}{2}-1\text{, }|u|\leq 1$$