Whakaoti i te n^2+2y^2+2x+2y+1=0

$${ n }^{ 2 } +2 { y }^{ 2 } +2x+2y+1 = 0$$

Whakaoti mō x

$$x=-\frac{n^{2}}{2}-y^{2}-y-\frac{1}{2}$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tangohia te n^{2} mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$2y^{2}+2x+2y+1=-n^{2}$$

Tangohia te 2y^{2} mai i ngā taha e rua.

$$2x+2y+1=-n^{2}-2y^{2}$$

Tangohia te 2y mai i ngā taha e rua.

$$2x+1=-n^{2}-2y^{2}-2y$$

Tangohia te 1 mai i ngā taha e rua.

$$2x=-n^{2}-2y^{2}-2y-1$$

He hanga arowhānui tō te whārite.

$$2x=-2y^{2}-2y-n^{2}-1$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x}{2}=\frac{-2y^{2}-2y-n^{2}-1}{2}$$

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

$$x=\frac{-2y^{2}-2y-n^{2}-1}{2}$$

Whakawehe -n^{2}-2y^{2}-2y-1 ki te 2.

$$x=-\frac{n^{2}}{2}-y^{2}-y-\frac{1}{2}$$

Whakaoti mō n (complex solution)

$$n=-\sqrt{-2x-2y^{2}-2y-1}$$

$$n=\sqrt{-2x-2y^{2}-2y-1}$$

Whakaoti mō n

$$n=\sqrt{-2x-2y^{2}-2y-1}$$

$$n=-\sqrt{-2x-2y^{2}-2y-1}\text{, }x\leq -y^{2}-y-\frac{1}{2}$$