Whakaoti i te b^2+3b-4

$$b ^ { 2 } + 3 b - 4$$

Tauwehe

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

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei b^{2}+pb+qb-4. Hei kimi p me q, whakaritea tētahi pūnaha kia whakaoti.

$$p+q=3 pq=1\left(-4\right)=-4$$

I te mea kua tōraro te pq, he tauaro ngā tohu o p me q. I te mea kua tōrunga te p+q, 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.

$$p=-1 q=4$$

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

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

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

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

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

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

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.

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

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.

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

Pūrua 3.

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

Whakareatia -4 ki te -4.

$$b=\frac{-3±\sqrt{9+16}}{2}$$

Tāpiri 9 ki te 16.

$$b=\frac{-3±\sqrt{25}}{2}$$

Tuhia te pūtakerua o te 25.

$$b=\frac{-3±5}{2}$$

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

$$b=\frac{2}{2}$$

Whakawehe 2 ki te 2.

$$b=1$$

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

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

Whakawehe -8 ki te 2.

$$b=-4$$

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}.

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

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

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

Aromātai

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