Whakaoti i te 18t^2-9t-5

$$18 t ^ { 2 } - 9 t - 5$$

Tauwehe

$$\left(6t-5\right)\left(3t+1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 18t^{2}+at+bt-5. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-9 ab=18\left(-5\right)=-90$$

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 -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Tātaihia te tapeke mō ia takirua.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -9.

$$a=-15 b=6$$

Tuhia anō te 18t^{2}-9t-5 hei \left(18t^{2}-15t\right)+\left(6t-5\right).

$$\left(18t^{2}-15t\right)+\left(6t-5\right)$$

Whakatauwehea atu 3t i te 18t^{2}-15t.

$$3t\left(6t-5\right)+6t-5$$

Whakatauwehea atu te kīanga pātahi 6t-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(6t-5\right)\left(3t+1\right)$$

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.

$$18t^{2}-9t-5=0$$

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.

$$t=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 18\left(-5\right)}}{2\times 18}$$

Pūrua -9.

$$t=\frac{-\left(-9\right)±\sqrt{81-4\times 18\left(-5\right)}}{2\times 18}$$

Whakareatia -4 ki te 18.

$$t=\frac{-\left(-9\right)±\sqrt{81-72\left(-5\right)}}{2\times 18}$$

Whakareatia -72 ki te -5.

$$t=\frac{-\left(-9\right)±\sqrt{81+360}}{2\times 18}$$

Tāpiri 81 ki te 360.

$$t=\frac{-\left(-9\right)±\sqrt{441}}{2\times 18}$$

Tuhia te pūtakerua o te 441.

$$t=\frac{-\left(-9\right)±21}{2\times 18}$$

Ko te tauaro o -9 ko 9.

$$t=\frac{9±21}{2\times 18}$$

Whakareatia 2 ki te 18.

$$t=\frac{9±21}{36}$$

Nā, me whakaoti te whārite t=\frac{9±21}{36} ina he tāpiri te ±. Tāpiri 9 ki te 21.

$$t=\frac{30}{36}$$

Whakahekea te hautanga \frac{30}{36} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

$$t=\frac{5}{6}$$

Nā, me whakaoti te whārite t=\frac{9±21}{36} ina he tango te ±. Tango 21 mai i 9.

$$t=-\frac{12}{36}$$

Whakahekea te hautanga \frac{-12}{36} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 12.

$$t=-\frac{1}{3}$$

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 \frac{5}{6} mō te x_{1} me te -\frac{1}{3} mō te x_{2}.

$$18t^{2}-9t-5=18\left(t-\frac{5}{6}\right)\left(t-\left(-\frac{1}{3}\right)\right)$$

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

$$18t^{2}-9t-5=18\left(t-\frac{5}{6}\right)\left(t+\frac{1}{3}\right)$$

Tango \frac{5}{6} mai i t mā te kimi i te tauraro pātahi me te tango i ngā taurunga, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$18t^{2}-9t-5=18\times \frac{6t-5}{6}\left(t+\frac{1}{3}\right)$$

Tāpiri \frac{1}{3} 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.

$$18t^{2}-9t-5=18\times \frac{6t-5}{6}\times \frac{3t+1}{3}$$

Whakareatia \frac{6t-5}{6} ki te \frac{3t+1}{3} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$18t^{2}-9t-5=18\times \frac{\left(6t-5\right)\left(3t+1\right)}{6\times 3}$$

Whakareatia 6 ki te 3.

$$18t^{2}-9t-5=18\times \frac{\left(6t-5\right)\left(3t+1\right)}{18}$$

Whakakorea atu te tauwehe pūnoa nui rawa 18 i roto i te 18 me te 18.

$$18t^{2}-9t-5=\left(6t-5\right)\left(3t+1\right)$$

Aromātai

$$\left(6t-5\right)\left(3t+1\right)$$