Whakaoti i te x^2+x+y^2=(x+2y)^2

$$x ^ { 2 } + x + y ^ { 2 } = ( x + 2 y ) ^ { 2 }$$

Whakaoti mō x

$$x=\frac{3y^{2}}{1-4y}$$

$$y\neq \frac{1}{4}$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+2y\right)^{2}.

$$x^{2}+x+y^{2}=x^{2}+4xy+4y^{2}$$

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

$$x^{2}+x+y^{2}-x^{2}=4xy+4y^{2}$$

Pahekotia te x^{2} me -x^{2}, ka 0.

$$x+y^{2}=4xy+4y^{2}$$

Tangohia te 4xy mai i ngā taha e rua.

$$x+y^{2}-4xy=4y^{2}$$

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

$$x-4xy=4y^{2}-y^{2}$$

Pahekotia te 4y^{2} me -y^{2}, ka 3y^{2}.

$$x-4xy=3y^{2}$$

Pahekotia ngā kīanga tau katoa e whai ana i te x.

$$\left(1-4y\right)x=3y^{2}$$

Whakawehea ngā taha e rua ki te -4y+1.

$$\frac{\left(1-4y\right)x}{1-4y}=\frac{3y^{2}}{1-4y}$$

Mā te whakawehe ki te -4y+1 ka wetekia te whakareanga ki te -4y+1.

$$x=\frac{3y^{2}}{1-4y}$$

Whakaoti mō y (complex solution)

$$y=\frac{-\sqrt{x\left(4x+3\right)}-2x}{3}$$

$$y=\frac{\sqrt{x\left(4x+3\right)}-2x}{3}$$

Whakaoti mō y

$$y=\frac{-\sqrt{x\left(4x+3\right)}-2x}{3}$$

$$y=\frac{\sqrt{x\left(4x+3\right)}-2x}{3}\text{, }x\leq -\frac{3}{4}\text{ or }x\geq 0$$