Whakaoti i te q(t)=t^2+t-20

$$q ( t ) = t ^ { 2 } + t - 20$$

Tauwehe

$$\left(t-4\right)\left(t+5\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei t^{2}+at+bt-20. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=1 ab=1\left(-20\right)=-20$$

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 -20.

$$-1,20 -2,10 -4,5$$

Tātaihia te tapeke mō ia takirua.

$$-1+20=19 -2+10=8 -4+5=1$$

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

$$a=-4 b=5$$

Tuhia anō te t^{2}+t-20 hei \left(t^{2}-4t\right)+\left(5t-20\right).

$$\left(t^{2}-4t\right)+\left(5t-20\right)$$

Tauwehea te t i te tuatahi me te 5 i te rōpū tuarua.

$$t\left(t-4\right)+5\left(t-4\right)$$

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

$$\left(t-4\right)\left(t+5\right)$$

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.

$$t^{2}+t-20=0$$

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.

$$t=\frac{-1±\sqrt{1^{2}-4\left(-20\right)}}{2}$$

Pūrua 1.

$$t=\frac{-1±\sqrt{1-4\left(-20\right)}}{2}$$

Whakareatia -4 ki te -20.

$$t=\frac{-1±\sqrt{1+80}}{2}$$

Tāpiri 1 ki te 80.

$$t=\frac{-1±\sqrt{81}}{2}$$

Tuhia te pūtakerua o te 81.

$$t=\frac{-1±9}{2}$$

Nā, me whakaoti te whārite t=\frac{-1±9}{2} ina he tāpiri te ±. Tāpiri -1 ki te 9.

$$t=\frac{8}{2}$$

Whakawehe 8 ki te 2.

$$t=4$$

Nā, me whakaoti te whārite t=\frac{-1±9}{2} ina he tango te ±. Tango 9 mai i -1.

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

Whakawehe -10 ki te 2.

$$t=-5$$

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 -5 mō te x_{2}.

$$t^{2}+t-20=\left(t-4\right)\left(t-\left(-5\right)\right)$$

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

$$t^{2}+t-20=\left(t-4\right)\left(t+5\right)$$

Aromātai

$$\left(t-4\right)\left(t+5\right)$$