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

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

Whakaoti mō x

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

$$y-1\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

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

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

Tangohia te 1 mai i ngā taha e rua.

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

Pūruatia ngā taha e rua o te whārite.

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

Me tango 2 mai i ngā taha e rua o te whārite.

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

Mā te tango i te 2 i a ia ake anō ka toe ko te 0.

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

Whakawehea ngā taha e rua ki te 2.

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

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

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

Whakawehe \left(y-1\right)^{2}-2 ki te 2.

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

Whakaoti mō x (complex solution)

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

$$y=1\text{ or }arg(1-y)\geq \pi$$

Whakaoti mō y (complex solution)

$$y=\sqrt{2\left(x+1\right)}+1$$

Whakaoti mō y

$$y=\sqrt{2\left(x+1\right)}+1$$

$$x\geq -1$$