Whakaoti i te 4t^2-13t-12

$$4 t ^ { 2 } - 13 t - 12$$

Tauwehe

$$\left(t-4\right)\left(4t+3\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 4t^{2}+at+bt-12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-13 ab=4\left(-12\right)=-48$$

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 -48.

$$1,-48 2,-24 3,-16 4,-12 6,-8$$

Tātaihia te tapeke mō ia takirua.

$$1-48=-47 2-24=-22 3-16=-13 4-12=-8 6-8=-2$$

Ko te otinga te takirua ka hoatu i te tapeke -13.

$$a=-16 b=3$$

Tuhia anō te 4t^{2}-13t-12 hei \left(4t^{2}-16t\right)+\left(3t-12\right).

$$\left(4t^{2}-16t\right)+\left(3t-12\right)$$

Tauwehea te 4t i te tuatahi me te 3 i te rōpū tuarua.

$$4t\left(t-4\right)+3\left(t-4\right)$$

Whakatauwehea atu te kīanga pātahi t-4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(t-4\right)\left(4t+3\right)$$

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.

$$4t^{2}-13t-12=0$$

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.

$$t=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 4\left(-12\right)}}{2\times 4}$$

Pūrua -13.

$$t=\frac{-\left(-13\right)±\sqrt{169-4\times 4\left(-12\right)}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$t=\frac{-\left(-13\right)±\sqrt{169-16\left(-12\right)}}{2\times 4}$$

Whakareatia -16 ki te -12.

$$t=\frac{-\left(-13\right)±\sqrt{169+192}}{2\times 4}$$

Tāpiri 169 ki te 192.

$$t=\frac{-\left(-13\right)±\sqrt{361}}{2\times 4}$$

Tuhia te pūtakerua o te 361.

$$t=\frac{-\left(-13\right)±19}{2\times 4}$$

Ko te tauaro o -13 ko 13.

$$t=\frac{13±19}{2\times 4}$$

Whakareatia 2 ki te 4.

$$t=\frac{13±19}{8}$$

Nā, me whakaoti te whārite t=\frac{13±19}{8} ina he tāpiri te ±. Tāpiri 13 ki te 19.

$$t=\frac{32}{8}$$

Whakawehe 32 ki te 8.

$$t=4$$

Nā, me whakaoti te whārite t=\frac{13±19}{8} ina he tango te ±. Tango 19 mai i 13.

$$t=-\frac{6}{8}$$

Whakahekea te hautanga \frac{-6}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$t=-\frac{3}{4}$$

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 4 mō te x_{1} me te -\frac{3}{4} mō te x_{2}.

$$4t^{2}-13t-12=4\left(t-4\right)\left(t-\left(-\frac{3}{4}\right)\right)$$

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

$$4t^{2}-13t-12=4\left(t-4\right)\left(t+\frac{3}{4}\right)$$

Tāpiri \frac{3}{4} ki te t 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.

$$4t^{2}-13t-12=4\left(t-4\right)\times \frac{4t+3}{4}$$

Whakakorea atu te tauwehe pūnoa nui rawa 4 i roto i te 4 me te 4.

$$4t^{2}-13t-12=\left(t-4\right)\left(4t+3\right)$$

Aromātai

$$\left(t-4\right)\left(4t+3\right)$$