Whakaoti i te t^2+3t-18

$$t ^ { 2 } + 3 t - 18$$

Tauwehe

$$\left(t-3\right)\left(t+6\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei t^{2}+at+bt-18. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=3 ab=1\left(-18\right)=-18$$

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 -18.

$$-1,18 -2,9 -3,6$$

Tātaihia te tapeke mō ia takirua.

$$-1+18=17 -2+9=7 -3+6=3$$

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

$$a=-3 b=6$$

Tuhia anō te t^{2}+3t-18 hei \left(t^{2}-3t\right)+\left(6t-18\right).

$$\left(t^{2}-3t\right)+\left(6t-18\right)$$

Tauwehea te t i te tuatahi me te 6 i te rōpū tuarua.

$$t\left(t-3\right)+6\left(t-3\right)$$

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

$$\left(t-3\right)\left(t+6\right)$$

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.

$$t^{2}+3t-18=0$$

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.

$$t=\frac{-3±\sqrt{3^{2}-4\left(-18\right)}}{2}$$

Pūrua 3.

$$t=\frac{-3±\sqrt{9-4\left(-18\right)}}{2}$$

Whakareatia -4 ki te -18.

$$t=\frac{-3±\sqrt{9+72}}{2}$$

Tāpiri 9 ki te 72.

$$t=\frac{-3±\sqrt{81}}{2}$$

Tuhia te pūtakerua o te 81.

$$t=\frac{-3±9}{2}$$

Nā, me whakaoti te whārite t=\frac{-3±9}{2} ina he tāpiri te ±. Tāpiri -3 ki te 9.

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

Whakawehe 6 ki te 2.

$$t=3$$

Nā, me whakaoti te whārite t=\frac{-3±9}{2} ina he tango te ±. Tango 9 mai i -3.

$$t=-\frac{12}{2}$$

Whakawehe -12 ki te 2.

$$t=-6$$

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 -6 mō te x_{2}.

$$t^{2}+3t-18=\left(t-3\right)\left(t-\left(-6\right)\right)$$

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

$$t^{2}+3t-18=\left(t-3\right)\left(t+6\right)$$

Aromātai

$$\left(t-3\right)\left(t+6\right)$$