Whakaoti i te y=sqrt{1/4x+2}

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

Whakaoti mō x

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

$$y\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

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

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

Pūruatia ngā taha e rua o te whārite.

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

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

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

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

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

Me whakarea ngā taha e rua ki te 4.

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

Mā te whakawehe ki te \frac{1}{4} ka wetekia te whakareanga ki te \frac{1}{4}.

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

Whakawehe y^{2}-2 ki te \frac{1}{4} mā te whakarea y^{2}-2 ki te tau huripoki o \frac{1}{4}.

$$x=4y^{2}-8$$

Whakaoti mō x (complex solution)

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

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

Whakaoti mō y (complex solution)

$$y=\frac{\sqrt{x+8}}{2}$$

Whakaoti mō y

$$y=\frac{\sqrt{x+8}}{2}$$

$$x\geq -8$$