Whakaoti i te y=sqrt{x+y}

$$y = \sqrt { x + y }$$

Whakaoti mō x

$$x=y\left(y-1\right)$$

$$y\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$\sqrt{x+y}=y$$

Pūruatia ngā taha e rua o te whārite.

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

Me tango y mai i ngā taha e rua o te whārite.

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

Mā te tango i te y i a ia ake anō ka toe ko te 0.

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

Tango y mai i y^{2}.

$$x=y\left(y-1\right)$$

Whakaoti mō x (complex solution)

$$x=y\left(y-1\right)$$

$$arg(y)<\pi \text{ or }y=0$$

Whakaoti mō y (complex solution)

$$\left\{\begin{matrix}y=\frac{-\sqrt{4x+1}+1}{2}\text{, }&arg(\frac{-\sqrt{4x+1}+1}{2})<\pi \text{ or }x=0\\y=\frac{\sqrt{4x+1}+1}{2}\text{, }&arg(\frac{\sqrt{4x+1}+1}{2})<\pi \end{matrix}\right.$$

Whakaoti mō y

$$\left\{\begin{matrix}y=\frac{\sqrt{4x+1}+1}{2}\text{, }&x\geq -\frac{1}{4}\\y=\frac{-\sqrt{4x+1}+1}{2}\text{, }&x\geq -\frac{1}{4}\text{ and }x\leq 0\end{matrix}\right.$$