Whakaoti i te x^2+3x-4.x^2+7x+12

$$x ^ { 2 } + 3 x - 4 . x ^ { 2 } + 7 x + 12$$

Tauwehe

$$\left(x-1\right)\left(x+4\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx-4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=3 ab=1\left(-4\right)=-4$$

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 -4.

$$-1,4 -2,2$$

Tātaihia te tapeke mō ia takirua.

$$-1+4=3 -2+2=0$$

Ko te otinga te takirua ka hoatu i te tapeke 3.

$$a=-1 b=4$$

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

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

Tauwehea te x i te tuatahi me te 4 i te rōpū tuarua.

$$x\left(x-1\right)+4\left(x-1\right)$$

Whakatauwehea atu te kīanga pātahi x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-1\right)\left(x+4\right)$$

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.

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

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.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-4\right)}}{2}$$

Pūrua 3.

$$x=\frac{-3±\sqrt{9-4\left(-4\right)}}{2}$$

Whakareatia -4 ki te -4.

$$x=\frac{-3±\sqrt{9+16}}{2}$$

Tāpiri 9 ki te 16.

$$x=\frac{-3±\sqrt{25}}{2}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-3±5}{2}$$

Nā, me whakaoti te whārite x=\frac{-3±5}{2} ina he tāpiri te ±. Tāpiri -3 ki te 5.

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

Whakawehe 2 ki te 2.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{-3±5}{2} ina he tango te ±. Tango 5 mai i -3.

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

Whakawehe -8 ki te 2.

$$x=-4$$

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 1 mō te x_{1} me te -4 mō te x_{2}.

$$x^{2}+3x-4=\left(x-1\right)\left(x-\left(-4\right)\right)$$

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

$$x^{2}+3x-4=\left(x-1\right)\left(x+4\right)$$

Aromātai

$$\left(x-1\right)\left(x+4\right)$$