Whakaoti i te 6t^2+t-12

$$6 t ^ { 2 } + t - 12$$

Tauwehe

$$\left(3t-4\right)\left(2t+3\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 6t^{2}+at+bt-12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=1 ab=6\left(-12\right)=-72$$

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 -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Tātaihia te tapeke mō ia takirua.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

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

$$a=-8 b=9$$

Tuhia anō te 6t^{2}+t-12 hei \left(6t^{2}-8t\right)+\left(9t-12\right).

$$\left(6t^{2}-8t\right)+\left(9t-12\right)$$

Tauwehea te 2t i te tuatahi me te 3 i te rōpū tuarua.

$$2t\left(3t-4\right)+3\left(3t-4\right)$$

Whakatauwehea atu te kīanga pātahi 3t-4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3t-4\right)\left(2t+3\right)$$

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.

$$6t^{2}+t-12=0$$

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.

$$t=\frac{-1±\sqrt{1^{2}-4\times 6\left(-12\right)}}{2\times 6}$$

Pūrua 1.

$$t=\frac{-1±\sqrt{1-4\times 6\left(-12\right)}}{2\times 6}$$

Whakareatia -4 ki te 6.

$$t=\frac{-1±\sqrt{1-24\left(-12\right)}}{2\times 6}$$

Whakareatia -24 ki te -12.

$$t=\frac{-1±\sqrt{1+288}}{2\times 6}$$

Tāpiri 1 ki te 288.

$$t=\frac{-1±\sqrt{289}}{2\times 6}$$

Tuhia te pūtakerua o te 289.

$$t=\frac{-1±17}{2\times 6}$$

Whakareatia 2 ki te 6.

$$t=\frac{-1±17}{12}$$

Nā, me whakaoti te whārite t=\frac{-1±17}{12} ina he tāpiri te ±. Tāpiri -1 ki te 17.

$$t=\frac{16}{12}$$

Whakahekea te hautanga \frac{16}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

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

Nā, me whakaoti te whārite t=\frac{-1±17}{12} ina he tango te ±. Tango 17 mai i -1.

$$t=-\frac{18}{12}$$

Whakahekea te hautanga \frac{-18}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

$$t=-\frac{3}{2}$$

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

$$6t^{2}+t-12=6\left(t-\frac{4}{3}\right)\left(t-\left(-\frac{3}{2}\right)\right)$$

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

$$6t^{2}+t-12=6\left(t-\frac{4}{3}\right)\left(t+\frac{3}{2}\right)$$

Tango \frac{4}{3} 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.

$$6t^{2}+t-12=6\times \frac{3t-4}{3}\left(t+\frac{3}{2}\right)$$

Tāpiri \frac{3}{2} 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.

$$6t^{2}+t-12=6\times \frac{3t-4}{3}\times \frac{2t+3}{2}$$

Whakareatia \frac{3t-4}{3} ki te \frac{2t+3}{2} 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.

$$6t^{2}+t-12=6\times \frac{\left(3t-4\right)\left(2t+3\right)}{3\times 2}$$

Whakareatia 3 ki te 2.

$$6t^{2}+t-12=6\times \frac{\left(3t-4\right)\left(2t+3\right)}{6}$$

Whakakorea atu te tauwehe pūnoa nui rawa 6 i roto i te 6 me te 6.

$$6t^{2}+t-12=\left(3t-4\right)\left(2t+3\right)$$

Aromātai

$$6t^{2}+t-12$$