Whakaoti i te x^2+2sqrt{2}kx+18=0

$${ x }^{ 2 } +2 \sqrt{ 2 } kx+18=0$$

Whakaoti mō k

$$k=-\frac{\sqrt{2}\left(x^{2}+18\right)}{4x}$$

$$x\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tangohia te x^{2} mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$2\sqrt{2}kx+18=-x^{2}$$

Tangohia te 18 mai i ngā taha e rua.

$$2\sqrt{2}kx=-x^{2}-18$$

He hanga arowhānui tō te whārite.

$$2\sqrt{2}xk=-x^{2}-18$$

Whakawehea ngā taha e rua ki te 2\sqrt{2}x.

$$\frac{2\sqrt{2}xk}{2\sqrt{2}x}=\frac{-x^{2}-18}{2\sqrt{2}x}$$

Mā te whakawehe ki te 2\sqrt{2}x ka wetekia te whakareanga ki te 2\sqrt{2}x.

$$k=\frac{-x^{2}-18}{2\sqrt{2}x}$$

Whakawehe -x^{2}-18 ki te 2\sqrt{2}x.

$$k=-\frac{\sqrt{2}\left(x^{2}+18\right)}{4x}$$

Whakaoti mō x (complex solution)

$$x=\sqrt{2}\left(\sqrt{k^{2}-9}-k\right)$$

$$x=\sqrt{2}\left(-\sqrt{k^{2}-9}-k\right)$$

Whakaoti mō x

$$x=\sqrt{2}\left(\sqrt{k^{2}-9}-k\right)$$

$$x=\sqrt{2}\left(-\sqrt{k^{2}-9}-k\right)\text{, }|k|\geq 3$$