Whakaoti i te 12=(1-3x)(1-3x)d+(1+3x)(1+3x)

$$12 = ( 1 - 3 x ) ( 1 - 3 x ) d + ( 1 + 3 x ) ( 1 + 3 x )$$

Whakaoti mō d

$$d=-\frac{9x^{2}+6x-11}{\left(1-3x\right)^{2}}$$

$$x\neq \frac{1}{3}$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakareatia te 1-3x ki te 1-3x, ka \left(1-3x\right)^{2}.

$$12=\left(1-3x\right)^{2}d+\left(1+3x\right)\left(1+3x\right)$$

Whakareatia te 1+3x ki te 1+3x, ka \left(1+3x\right)^{2}.

$$12=\left(1-3x\right)^{2}d+\left(1+3x\right)^{2}$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(1-3x\right)^{2}.

$$12=\left(1-6x+9x^{2}\right)d+\left(1+3x\right)^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te 1-6x+9x^{2} ki te d.

$$12=d-6xd+9x^{2}d+\left(1+3x\right)^{2}$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(1+3x\right)^{2}.

$$12=d-6xd+9x^{2}d+1+6x+9x^{2}$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$d-6xd+9x^{2}d+1+6x+9x^{2}=12$$

Tangohia te 1 mai i ngā taha e rua.

$$d-6xd+9x^{2}d+6x+9x^{2}=12-1$$

Tangohia te 1 i te 12, ka 11.

$$d-6xd+9x^{2}d+6x+9x^{2}=11$$

Tangohia te 6x mai i ngā taha e rua.

$$d-6xd+9x^{2}d+9x^{2}=11-6x$$

Tangohia te 9x^{2} mai i ngā taha e rua.

$$d-6xd+9x^{2}d=11-6x-9x^{2}$$

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

$$\left(1-6x+9x^{2}\right)d=11-6x-9x^{2}$$

He hanga arowhānui tō te whārite.

$$\left(9x^{2}-6x+1\right)d=11-6x-9x^{2}$$

Whakawehea ngā taha e rua ki te 1-6x+9x^{2}.

$$\frac{\left(9x^{2}-6x+1\right)d}{9x^{2}-6x+1}=\frac{11-6x-9x^{2}}{9x^{2}-6x+1}$$

Mā te whakawehe ki te 1-6x+9x^{2} ka wetekia te whakareanga ki te 1-6x+9x^{2}.

$$d=\frac{11-6x-9x^{2}}{9x^{2}-6x+1}$$

Whakawehe 11-6x-9x^{2} ki te 1-6x+9x^{2}.

$$d=\frac{11-6x-9x^{2}}{\left(3x-1\right)^{2}}$$

Whakaoti mō x (complex solution)

$$\left\{\begin{matrix}x=-\frac{-d+2\sqrt{2d+3}+1}{3\left(d+1\right)}\text{; }x=-\frac{-d-2\sqrt{2d+3}+1}{3\left(d+1\right)}\text{, }&d\neq -1\\x=1\text{, }&d=-1\end{matrix}\right.$$

Whakaoti mō x

$$\left\{\begin{matrix}x=-\frac{-d+2\sqrt{2d+3}+1}{3\left(d+1\right)}\text{; }x=-\frac{-d-2\sqrt{2d+3}+1}{3\left(d+1\right)}\text{, }&d\neq -1\text{ and }d\geq -\frac{3}{2}\\x=1\text{, }&d=-1\end{matrix}\right.$$