Whakaoti i te 3x^2-ksqrt{3}x+4=0

$$3 { x }^{ 2 } -k \sqrt{ 3 } x+4=0$$

Whakaoti mō k

$$k=\sqrt{3}x+\frac{4\sqrt{3}}{3x}$$

$$x\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tangohia te 4 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$3x^{2}-k\sqrt{3}x=-4$$

Tangohia te 3x^{2} mai i ngā taha e rua.

$$-k\sqrt{3}x=-4-3x^{2}$$

He hanga arowhānui tō te whārite.

$$\left(-\sqrt{3}x\right)k=-3x^{2}-4$$

Whakawehea ngā taha e rua ki te -\sqrt{3}x.

$$\frac{\left(-\sqrt{3}x\right)k}{-\sqrt{3}x}=\frac{-3x^{2}-4}{-\sqrt{3}x}$$

Mā te whakawehe ki te -\sqrt{3}x ka wetekia te whakareanga ki te -\sqrt{3}x.

$$k=\frac{-3x^{2}-4}{-\sqrt{3}x}$$

Whakawehe -4-3x^{2} ki te -\sqrt{3}x.

$$k=\sqrt{3}x+\frac{4\sqrt{3}}{3x}$$

Whakaoti mō x (complex solution)

$$x=\frac{\sqrt{3}\left(\sqrt{k^{2}-16}+k\right)}{6}$$

$$x=-\frac{\sqrt{3}\left(\sqrt{k^{2}-16}-k\right)}{6}$$

Whakaoti mō x

$$x=-\frac{\sqrt{3}\left(\sqrt{k^{2}-16}-k\right)}{6}$$

$$x=\frac{\sqrt{3}\left(\sqrt{k^{2}-16}+k\right)}{6}\text{, }|k|\geq 4$$