Whakaoti i te kx^2+left(2k+1right)x+k-1=0

$$k { x }^{ 2 } + \left( 2k+1 \right) x+k-1 = 0$$

Whakaoti mō k

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

$$x\neq -1$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te 2k+1 ki te x.

$$kx^{2}+2kx+x+k-1=0$$

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

$$kx^{2}+2kx+k-1=-x$$

Me tāpiri te 1 ki ngā taha e rua.

$$kx^{2}+2kx+k=-x+1$$

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

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

He hanga arowhānui tō te whārite.

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

Whakawehea ngā taha e rua ki te x^{2}+2x+1.

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

Mā te whakawehe ki te x^{2}+2x+1 ka wetekia te whakareanga ki te x^{2}+2x+1.

$$k=\frac{1-x}{x^{2}+2x+1}$$

Whakawehe -x+1 ki te x^{2}+2x+1.

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

Whakaoti mō x (complex solution)

$$\left\{\begin{matrix}x=\frac{-2k+\sqrt{8k+1}-1}{2k}\text{; }x=-\frac{2k+\sqrt{8k+1}+1}{2k}\text{, }&k\neq 0\\x=1\text{, }&k=0\end{matrix}\right.$$

Whakaoti mō x

$$\left\{\begin{matrix}x=\frac{-2k+\sqrt{8k+1}-1}{2k}\text{; }x=-\frac{2k+\sqrt{8k+1}+1}{2k}\text{, }&k\neq 0\text{ and }k\geq -\frac{1}{8}\\x=1\text{, }&k=0\end{matrix}\right.$$