Whakaoti i te x^2-2(m-1)x+2m+1=0

$$x ^ { 2 } - 2 ( m - 1 ) x + 2 m + 1 = 0$$

Whakaoti mō m

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

$$x\neq 1$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tangohia te 1 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

Whakamahia te āhuatanga tohatoha hei whakarea te -2 ki te m-1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te -2m+2 ki te x.

$$x^{2}-2mx+2x+2m=-1$$

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

$$-2mx+2x+2m=-1-x^{2}$$

Tangohia te 2x mai i ngā taha e rua.

$$-2mx+2m=-1-x^{2}-2x$$

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

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

He hanga arowhānui tō te whārite.

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

Whakawehea ngā taha e rua ki te -2x+2.

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

Mā te whakawehe ki te -2x+2 ka wetekia te whakareanga ki te -2x+2.

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

Whakawehe -\left(x+1\right)^{2} ki te -2x+2.

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

Whakaoti mō x (complex solution)

$$x=\sqrt{m\left(m-4\right)}+m-1$$

$$x=-\sqrt{m\left(m-4\right)}+m-1$$

Whakaoti mō x

$$x=\sqrt{m\left(m-4\right)}+m-1$$

$$x=-\sqrt{m\left(m-4\right)}+m-1\text{, }m\geq 4\text{ or }m\leq 0$$