Whakaoti i te V=sqrt{4x-2}

$$V = \sqrt { 4 x - 2 }$$

Whakaoti mō x

$$x=\frac{V^{2}+2}{4}$$

$$V\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

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

$$\sqrt{4x-2}=V$$

Pūruatia ngā taha e rua o te whārite.

$$4x-2=V^{2}$$

Me tāpiri 2 ki ngā taha e rua o te whārite.

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

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

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

Tango -2 mai i V^{2}.

$$4x=V^{2}+2$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x}{4}=\frac{V^{2}+2}{4}$$

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

$$x=\frac{V^{2}+2}{4}$$

Whakawehe V^{2}+2 ki te 4.

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

Whakaoti mō x (complex solution)

$$x=\frac{V^{2}+2}{4}$$

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

Whakaoti mō V

$$V=\sqrt{4x-2}$$

$$x\geq \frac{1}{2}$$