Whakaoti i te 9p^2-8p-1

$$9 p ^ { 2 } - 8 p - 1$$

Tauwehe

$$\left(p-1\right)\left(9p+1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 9p^{2}+ap+bp-1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-8 ab=9\left(-1\right)=-9$$

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 -9.

$$1,-9 3,-3$$

Tātaihia te tapeke mō ia takirua.

$$1-9=-8 3-3=0$$

Ko te otinga te takirua ka hoatu i te tapeke -8.

$$a=-9 b=1$$

Tuhia anō te 9p^{2}-8p-1 hei \left(9p^{2}-9p\right)+\left(p-1\right).

$$\left(9p^{2}-9p\right)+\left(p-1\right)$$

Whakatauwehea atu 9p i te 9p^{2}-9p.

$$9p\left(p-1\right)+p-1$$

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

$$\left(p-1\right)\left(9p+1\right)$$

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.

$$9p^{2}-8p-1=0$$

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.

$$p=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 9\left(-1\right)}}{2\times 9}$$

Pūrua -8.

$$p=\frac{-\left(-8\right)±\sqrt{64-4\times 9\left(-1\right)}}{2\times 9}$$

Whakareatia -4 ki te 9.

$$p=\frac{-\left(-8\right)±\sqrt{64-36\left(-1\right)}}{2\times 9}$$

Whakareatia -36 ki te -1.

$$p=\frac{-\left(-8\right)±\sqrt{64+36}}{2\times 9}$$

Tāpiri 64 ki te 36.

$$p=\frac{-\left(-8\right)±\sqrt{100}}{2\times 9}$$

Tuhia te pūtakerua o te 100.

$$p=\frac{-\left(-8\right)±10}{2\times 9}$$

Ko te tauaro o -8 ko 8.

$$p=\frac{8±10}{2\times 9}$$

Whakareatia 2 ki te 9.

$$p=\frac{8±10}{18}$$

Nā, me whakaoti te whārite p=\frac{8±10}{18} ina he tāpiri te ±. Tāpiri 8 ki te 10.

$$p=\frac{18}{18}$$

Whakawehe 18 ki te 18.

$$p=1$$

Nā, me whakaoti te whārite p=\frac{8±10}{18} ina he tango te ±. Tango 10 mai i 8.

$$p=-\frac{2}{18}$$

Whakahekea te hautanga \frac{-2}{18} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$p=-\frac{1}{9}$$

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 -\frac{1}{9} mō te x_{2}.

$$9p^{2}-8p-1=9\left(p-1\right)\left(p-\left(-\frac{1}{9}\right)\right)$$

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

$$9p^{2}-8p-1=9\left(p-1\right)\left(p+\frac{1}{9}\right)$$

Tāpiri \frac{1}{9} ki te p mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$9p^{2}-8p-1=9\left(p-1\right)\times \frac{9p+1}{9}$$

Whakakorea atu te tauwehe pūnoa nui rawa 9 i roto i te 9 me te 9.

$$9p^{2}-8p-1=\left(p-1\right)\left(9p+1\right)$$

Aromātai

$$\left(p-1\right)\left(9p+1\right)$$