Whakaoti i te 3r^2+r-14

$$3 r ^ { 2 } + r - 14$$

Tauwehe

$$\left(r-2\right)\left(3r+7\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3r^{2}+ar+br-14. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=1 ab=3\left(-14\right)=-42$$

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 -42.

$$-1,42 -2,21 -3,14 -6,7$$

Tātaihia te tapeke mō ia takirua.

$$-1+42=41 -2+21=19 -3+14=11 -6+7=1$$

Ko te otinga te takirua ka hoatu i te tapeke 1.

$$a=-6 b=7$$

Tuhia anō te 3r^{2}+r-14 hei \left(3r^{2}-6r\right)+\left(7r-14\right).

$$\left(3r^{2}-6r\right)+\left(7r-14\right)$$

Tauwehea te 3r i te tuatahi me te 7 i te rōpū tuarua.

$$3r\left(r-2\right)+7\left(r-2\right)$$

Whakatauwehea atu te kīanga pātahi r-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(r-2\right)\left(3r+7\right)$$

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.

$$3r^{2}+r-14=0$$

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.

$$r=\frac{-1±\sqrt{1^{2}-4\times 3\left(-14\right)}}{2\times 3}$$

Pūrua 1.

$$r=\frac{-1±\sqrt{1-4\times 3\left(-14\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$r=\frac{-1±\sqrt{1-12\left(-14\right)}}{2\times 3}$$

Whakareatia -12 ki te -14.

$$r=\frac{-1±\sqrt{1+168}}{2\times 3}$$

Tāpiri 1 ki te 168.

$$r=\frac{-1±\sqrt{169}}{2\times 3}$$

Tuhia te pūtakerua o te 169.

$$r=\frac{-1±13}{2\times 3}$$

Whakareatia 2 ki te 3.

$$r=\frac{-1±13}{6}$$

Nā, me whakaoti te whārite r=\frac{-1±13}{6} ina he tāpiri te ±. Tāpiri -1 ki te 13.

$$r=\frac{12}{6}$$

Whakawehe 12 ki te 6.

$$r=2$$

Nā, me whakaoti te whārite r=\frac{-1±13}{6} ina he tango te ±. Tango 13 mai i -1.

$$r=-\frac{14}{6}$$

Whakahekea te hautanga \frac{-14}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$r=-\frac{7}{3}$$

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 2 mō te x_{1} me te -\frac{7}{3} mō te x_{2}.

$$3r^{2}+r-14=3\left(r-2\right)\left(r-\left(-\frac{7}{3}\right)\right)$$

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

$$3r^{2}+r-14=3\left(r-2\right)\left(r+\frac{7}{3}\right)$$

Tāpiri \frac{7}{3} ki te r 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.

$$3r^{2}+r-14=3\left(r-2\right)\times \frac{3r+7}{3}$$

Whakakorea atu te tauwehe pūnoa nui rawa 3 i roto i te 3 me te 3.

$$3r^{2}+r-14=\left(r-2\right)\left(3r+7\right)$$

Aromātai

$$\left(r-2\right)\left(3r+7\right)$$