Whakaoti i te x=sqrt{3y-1/2}

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

Whakaoti mō y

$$y=\frac{x^{2}}{3}+\frac{1}{6}$$

$$x\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

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

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

Pūruatia ngā taha e rua o te whārite.

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

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

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

Mā te tango i te -\frac{1}{2} i a ia ake anō ka toe ko te 0.

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

Tango -\frac{1}{2} mai i x^{2}.

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

Whakawehea ngā taha e rua ki te 3.

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

Mā te whakawehe ki te 3 ka wetekia te whakareanga ki te 3.

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

Whakawehe x^{2}+\frac{1}{2} ki te 3.

$$y=\frac{x^{2}}{3}+\frac{1}{6}$$

Whakaoti mō x (complex solution)

$$x=\frac{\sqrt{12y-2}}{2}$$

Whakaoti mō y (complex solution)

$$y=\frac{x^{2}}{3}+\frac{1}{6}$$

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

Whakaoti mō x

$$x=\frac{\sqrt{12y-2}}{2}$$

$$y\geq \frac{1}{6}$$