Whakaoti i te y=2sqrt{x+1}

$$y = 2 \sqrt { x + 1 }$$

Whakaoti mō x

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

$$y\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

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

$$2\sqrt{x+1}=y$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2\sqrt{x+1}}{2}=\frac{y}{2}$$

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

$$\sqrt{x+1}=\frac{y}{2}$$

Pūruatia ngā taha e rua o te whārite.

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

Me tango 1 mai i ngā taha e rua o te whārite.

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

Mā te tango i te 1 i a ia ake anō ka toe ko te 0.

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

Whakaoti mō x (complex solution)

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

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

Whakaoti mō y

$$y=2\sqrt{x+1}$$

$$x\geq -1$$