Whakaoti i te r^2-3r-130

$$r ^ { 2 } - 3 r - 130$$

Tauwehe

$$\left(r-13\right)\left(r+10\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei r^{2}+ar+br-130. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-3 ab=1\left(-130\right)=-130$$

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 -130.

$$1,-130 2,-65 5,-26 10,-13$$

Tātaihia te tapeke mō ia takirua.

$$1-130=-129 2-65=-63 5-26=-21 10-13=-3$$

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

$$a=-13 b=10$$

Tuhia anō te r^{2}-3r-130 hei \left(r^{2}-13r\right)+\left(10r-130\right).

$$\left(r^{2}-13r\right)+\left(10r-130\right)$$

Tauwehea te r i te tuatahi me te 10 i te rōpū tuarua.

$$r\left(r-13\right)+10\left(r-13\right)$$

Whakatauwehea atu te kīanga pātahi r-13 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(r-13\right)\left(r+10\right)$$

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.

$$r^{2}-3r-130=0$$

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.

$$r=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-130\right)}}{2}$$

Pūrua -3.

$$r=\frac{-\left(-3\right)±\sqrt{9-4\left(-130\right)}}{2}$$

Whakareatia -4 ki te -130.

$$r=\frac{-\left(-3\right)±\sqrt{9+520}}{2}$$

Tāpiri 9 ki te 520.

$$r=\frac{-\left(-3\right)±\sqrt{529}}{2}$$

Tuhia te pūtakerua o te 529.

$$r=\frac{-\left(-3\right)±23}{2}$$

Ko te tauaro o -3 ko 3.

$$r=\frac{3±23}{2}$$

Nā, me whakaoti te whārite r=\frac{3±23}{2} ina he tāpiri te ±. Tāpiri 3 ki te 23.

$$r=\frac{26}{2}$$

Whakawehe 26 ki te 2.

$$r=13$$

Nā, me whakaoti te whārite r=\frac{3±23}{2} ina he tango te ±. Tango 23 mai i 3.

$$r=-\frac{20}{2}$$

Whakawehe -20 ki te 2.

$$r=-10$$

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 13 mō te x_{1} me te -10 mō te x_{2}.

$$r^{2}-3r-130=\left(r-13\right)\left(r-\left(-10\right)\right)$$

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

$$r^{2}-3r-130=\left(r-13\right)\left(r+10\right)$$

Aromātai

$$\left(r-13\right)\left(r+10\right)$$