Whakaoti i te left(k-2right)x^2-x-4k=0

$$\left( k-2 \right) { x }^{ 2 } -x-4k=0$$

Whakaoti mō k (complex solution)

$$k=\frac{x\left(2x+1\right)}{x^{2}-4}$$

$$x\neq -2\text{ and }x\neq 2$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te k-2 ki te x^{2}.

$$kx^{2}-2x^{2}-x-4k=0$$

Me tāpiri te 2x^{2} ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$kx^{2}-x-4k=2x^{2}$$

Me tāpiri te x ki ngā taha e rua.

$$kx^{2}-4k=2x^{2}+x$$

Pahekotia ngā kīanga tau katoa e whai ana i te k.

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

Whakawehea ngā taha e rua ki te x^{2}-4.

$$\frac{\left(x^{2}-4\right)k}{x^{2}-4}=\frac{x\left(2x+1\right)}{x^{2}-4}$$

Mā te whakawehe ki te x^{2}-4 ka wetekia te whakareanga ki te x^{2}-4.

$$k=\frac{x\left(2x+1\right)}{x^{2}-4}$$

Whakaoti mō k

$$k=\frac{x\left(2x+1\right)}{x^{2}-4}$$

$$|x|\neq 2$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te k-2 ki te x^{2}.

$$kx^{2}-2x^{2}-x-4k=0$$

Me tāpiri te 2x^{2} ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$kx^{2}-x-4k=2x^{2}$$

Me tāpiri te x ki ngā taha e rua.

$$kx^{2}-4k=2x^{2}+x$$

Pahekotia ngā kīanga tau katoa e whai ana i te k.

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

Whakawehea ngā taha e rua ki te x^{2}-4.

$$\frac{\left(x^{2}-4\right)k}{x^{2}-4}=\frac{x\left(2x+1\right)}{x^{2}-4}$$

Mā te whakawehe ki te x^{2}-4 ka wetekia te whakareanga ki te x^{2}-4.

$$k=\frac{x\left(2x+1\right)}{x^{2}-4}$$

Whakaoti mō x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{16k^{2}-32k+1}+1}{2\left(k-2\right)}\text{; }x=\frac{-\sqrt{16k^{2}-32k+1}+1}{2\left(k-2\right)}\text{, }&k\neq 2\\x=-8\text{, }&k=2\end{matrix}\right.$$

Whakaoti mō x

$$\left\{\begin{matrix}x=\frac{\sqrt{16k^{2}-32k+1}+1}{2\left(k-2\right)}\text{; }x=\frac{-\sqrt{16k^{2}-32k+1}+1}{2\left(k-2\right)}\text{, }&k\leq -\frac{\sqrt{15}}{4}+1\text{ or }\left(k\neq 2\text{ and }k\geq \frac{\sqrt{15}}{4}+1\right)\\x=-8\text{, }&k=2\end{matrix}\right.$$