Whakaoti i te b^2-2b-3

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

Tauwehe

$$\left(b-3\right)\left(b+1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei b^{2}+pb+qb-3. Hei kimi p me q, whakaritea tētahi pūnaha kia whakaoti.

$$p+q=-2 pq=1\left(-3\right)=-3$$

I te mea kua tōraro te pq, he tauaro ngā tohu o p me q. I te mea kua tōraro te p+q, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Ko te takirua anake pērā ko te otinga pūnaha.

$$p=-3 q=1$$

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

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

Whakatauwehea atu b i te b^{2}-3b.

$$b\left(b-3\right)+b-3$$

Whakatauwehea atu te kīanga pātahi b-3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(b-3\right)\left(b+1\right)$$

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.

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

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.

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

Pūrua -2.

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

Whakareatia -4 ki te -3.

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

Tāpiri 4 ki te 12.

$$b=\frac{-\left(-2\right)±\sqrt{16}}{2}$$

Tuhia te pūtakerua o te 16.

$$b=\frac{-\left(-2\right)±4}{2}$$

Ko te tauaro o -2 ko 2.

$$b=\frac{2±4}{2}$$

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

$$b=\frac{6}{2}$$

Whakawehe 6 ki te 2.

$$b=3$$

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

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

Whakawehe -2 ki te 2.

$$b=-1$$

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 3 mō te x_{1} me te -1 mō te x_{2}.

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

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

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

Aromātai

$$\left(b-3\right)\left(b+1\right)$$