Whakaoti i te x^2-x-12

$$x ^ { 2 } - x - 12$$

Tauwehe

$$\left(x-4\right)\left(x+3\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx-12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=1\left(-12\right)=-12$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -12.

$$1,-12 2,-6 3,-4$$

Tātaihia te tapeke mō ia takirua.

$$1-12=-11 2-6=-4 3-4=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -1.

$$a=-4 b=3$$

Tuhia anō te x^{2}-x-12 hei \left(x^{2}-4x\right)+\left(3x-12\right).

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

Tauwehea te x i te tuatahi me te 3 i te rōpū tuarua.

$$x\left(x-4\right)+3\left(x-4\right)$$

Whakatauwehea atu te kīanga pātahi x-4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-4\right)\left(x+3\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Ka taea te huamaha pūrua te tauwehe mā te whakamahi i te huringa ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), ina ko x_{1} me x_{2} ngā otinga o te whārite pūrua ax^{2}+bx+c=0.

$$x^{2}-x-12=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-12\right)}}{2}$$

Whakareatia -4 ki te -12.

$$x=\frac{-\left(-1\right)±\sqrt{1+48}}{2}$$

Tāpiri 1 ki te 48.

$$x=\frac{-\left(-1\right)±\sqrt{49}}{2}$$

Tuhia te pūtakerua o te 49.

$$x=\frac{-\left(-1\right)±7}{2}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±7}{2}$$

Nā, me whakaoti te whārite x=\frac{1±7}{2} ina he tāpiri te ±. Tāpiri 1 ki te 7.

$$x=\frac{8}{2}$$

Whakawehe 8 ki te 2.

$$x=4$$

Nā, me whakaoti te whārite x=\frac{1±7}{2} ina he tango te ±. Tango 7 mai i 1.

$$x=-\frac{6}{2}$$

Whakawehe -6 ki te 2.

$$x=-3$$

Tauwehea te kīanga taketake mā te whakamahi i te ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Me whakakapi te 4 mō te x_{1} me te -3 mō te x_{2}.

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

Whakamāmātia ngā kīanga katoa o te āhua p-\left(-q\right) ki te p+q.

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

Aromātai

$$\left(x-4\right)\left(x+3\right)$$