Whakaoti i te x^2+4x-12

$${ x }^{ 2 } +4x-12$$

Tauwehe

$$\left(x-2\right)\left(x+6\right)$$

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.

$$a+b=4 ab=1\left(-12\right)=-12$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. 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 4.

$$a=-2 b=6$$

Tuhia anō te x^{2}+4x-12 hei \left(x^{2}-2x\right)+\left(6x-12\right).

$$\left(x^{2}-2x\right)+\left(6x-12\right)$$

Tauwehea te x i te tuatahi me te 6 i te rōpū tuarua.

$$x\left(x-2\right)+6\left(x-2\right)$$

Whakatauwehea atu te kīanga pātahi x-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-2\right)\left(x+6\right)$$

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.

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

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.

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

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\left(-12\right)}}{2}$$

Whakareatia -4 ki te -12.

$$x=\frac{-4±\sqrt{16+48}}{2}$$

Tāpiri 16 ki te 48.

$$x=\frac{-4±\sqrt{64}}{2}$$

Tuhia te pūtakerua o te 64.

$$x=\frac{-4±8}{2}$$

Nā, me whakaoti te whārite x=\frac{-4±8}{2} ina he tāpiri te ±. Tāpiri -4 ki te 8.

$$x=\frac{4}{2}$$

Whakawehe 4 ki te 2.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{-4±8}{2} ina he tango te ±. Tango 8 mai i -4.

$$x=-\frac{12}{2}$$

Whakawehe -12 ki te 2.

$$x=-6$$

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 2 mō te x_{1} me te -6 mō te x_{2}.

$$x^{2}+4x-12=\left(x-2\right)\left(x-\left(-6\right)\right)$$

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

$$x^{2}+4x-12=\left(x-2\right)\left(x+6\right)$$

Aromātai

$$\left(x-2\right)\left(x+6\right)$$