Whakaoti i te x^2+(3a-4)x-12a=0

$$x ^ { 2 } + ( 3 a - 4 ) x - 12 a = 0$$

Whakaoti mō a (complex solution)

$$\left\{\begin{matrix}\\a=-\frac{x}{3}\text{, }&\text{unconditionally}\\a\in \mathrm{C}\text{, }&x=4\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te 3a-4 ki te x.

$$x^{2}+3ax-4x-12a=0$$

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

$$3ax-4x-12a=-x^{2}$$

Me tāpiri te 4x ki ngā taha e rua.

$$3ax-12a=-x^{2}+4x$$

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

$$\left(3x-12\right)a=-x^{2}+4x$$

He hanga arowhānui tō te whārite.

$$\left(3x-12\right)a=4x-x^{2}$$

Whakawehea ngā taha e rua ki te 3x-12.

$$\frac{\left(3x-12\right)a}{3x-12}=\frac{x\left(4-x\right)}{3x-12}$$

Mā te whakawehe ki te 3x-12 ka wetekia te whakareanga ki te 3x-12.

$$a=\frac{x\left(4-x\right)}{3x-12}$$

Whakawehe x\left(4-x\right) ki te 3x-12.

$$a=-\frac{x}{3}$$

Whakaoti mō a

$$\left\{\begin{matrix}\\a=-\frac{x}{3}\text{, }&\text{unconditionally}\\a\in \mathrm{R}\text{, }&x=4\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te 3a-4 ki te x.

$$x^{2}+3ax-4x-12a=0$$

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

$$3ax-4x-12a=-x^{2}$$

Me tāpiri te 4x ki ngā taha e rua.

$$3ax-12a=-x^{2}+4x$$

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

$$\left(3x-12\right)a=-x^{2}+4x$$

He hanga arowhānui tō te whārite.

$$\left(3x-12\right)a=4x-x^{2}$$

Whakawehea ngā taha e rua ki te 3x-12.

$$\frac{\left(3x-12\right)a}{3x-12}=\frac{x\left(4-x\right)}{3x-12}$$

Mā te whakawehe ki te 3x-12 ka wetekia te whakareanga ki te 3x-12.

$$a=\frac{x\left(4-x\right)}{3x-12}$$

Whakawehe x\left(4-x\right) ki te 3x-12.

$$a=-\frac{x}{3}$$

Whakaoti mō x

$$x=-3a$$

$$x=4$$