Whakaoti i te v^2=4+8x

$$v ^ { 2 } = 4 + 8 x$$

Whakaoti mō x

$$x=\frac{v^{2}-4}{8}$$

Ngā Upane mō te Whakaoti Whārite Rārangi

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

$$4+8x=v^{2}$$

Tangohia te 4 mai i ngā taha e rua.

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

Whakawehea ngā taha e rua ki te 8.

$$\frac{8x}{8}=\frac{v^{2}-4}{8}$$

Mā te whakawehe ki te 8 ka wetekia te whakareanga ki te 8.

$$x=\frac{v^{2}-4}{8}$$

Whakawehe v^{2}-4 ki te 8.

$$x=\frac{v^{2}}{8}-\frac{1}{2}$$

Whakaoti mō v (complex solution)

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

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

Whakaoti mō v

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

$$v=-2\sqrt{2x+1}\text{, }x\geq -\frac{1}{2}$$