Whakaoti i te (t^2-3t-10)

$$( t ^ { 2 } - 3 t - 10 )$$

Tauwehe

$$\left(t-5\right)\left(t+2\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei t^{2}+at+bt-10. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-3 ab=1\left(-10\right)=-10$$

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 -10.

$$1,-10 2,-5$$

Tātaihia te tapeke mō ia takirua.

$$1-10=-9 2-5=-3$$

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

$$a=-5 b=2$$

Tuhia anō te t^{2}-3t-10 hei \left(t^{2}-5t\right)+\left(2t-10\right).

$$\left(t^{2}-5t\right)+\left(2t-10\right)$$

Tauwehea te t i te tuatahi me te 2 i te rōpū tuarua.

$$t\left(t-5\right)+2\left(t-5\right)$$

Whakatauwehea atu te kīanga pātahi t-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(t-5\right)\left(t+2\right)$$

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.

$$t^{2}-3t-10=0$$

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.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-10\right)}}{2}$$

Pūrua -3.

$$t=\frac{-\left(-3\right)±\sqrt{9-4\left(-10\right)}}{2}$$

Whakareatia -4 ki te -10.

$$t=\frac{-\left(-3\right)±\sqrt{9+40}}{2}$$

Tāpiri 9 ki te 40.

$$t=\frac{-\left(-3\right)±\sqrt{49}}{2}$$

Tuhia te pūtakerua o te 49.

$$t=\frac{-\left(-3\right)±7}{2}$$

Ko te tauaro o -3 ko 3.

$$t=\frac{3±7}{2}$$

Nā, me whakaoti te whārite t=\frac{3±7}{2} ina he tāpiri te ±. Tāpiri 3 ki te 7.

$$t=\frac{10}{2}$$

Whakawehe 10 ki te 2.

$$t=5$$

Nā, me whakaoti te whārite t=\frac{3±7}{2} ina he tango te ±. Tango 7 mai i 3.

$$t=-\frac{4}{2}$$

Whakawehe -4 ki te 2.

$$t=-2$$

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 5 mō te x_{1} me te -2 mō te x_{2}.

$$t^{2}-3t-10=\left(t-5\right)\left(t-\left(-2\right)\right)$$

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

$$t^{2}-3t-10=\left(t-5\right)\left(t+2\right)$$

Aromātai

$$\left(t-5\right)\left(t+2\right)$$