Whakaoti i te 2a^2+5a-12

$$2 a ^ { 2 } + 5 a - 12$$

Tauwehe

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

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 2a^{2}+pa+qa-12. Hei kimi p me q, whakaritea tētahi pūnaha kia whakaoti.

$$p+q=5 pq=2\left(-12\right)=-24$$

I te mea kua tōraro te pq, he tauaro ngā tohu o p me q. I te mea kua tōrunga te p+q, 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 -24.

$$-1,24 -2,12 -3,8 -4,6$$

Tātaihia te tapeke mō ia takirua.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

Ko te otinga te takirua ka hoatu i te tapeke 5.

$$p=-3 q=8$$

Tuhia anō te 2a^{2}+5a-12 hei \left(2a^{2}-3a\right)+\left(8a-12\right).

$$\left(2a^{2}-3a\right)+\left(8a-12\right)$$

Tauwehea te a i te tuatahi me te 4 i te rōpū tuarua.

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

Whakatauwehea atu te kīanga pātahi 2a-3 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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.

$$2a^{2}+5a-12=0$$

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.

$$a=\frac{-5±\sqrt{5^{2}-4\times 2\left(-12\right)}}{2\times 2}$$

Pūrua 5.

$$a=\frac{-5±\sqrt{25-4\times 2\left(-12\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$a=\frac{-5±\sqrt{25-8\left(-12\right)}}{2\times 2}$$

Whakareatia -8 ki te -12.

$$a=\frac{-5±\sqrt{25+96}}{2\times 2}$$

Tāpiri 25 ki te 96.

$$a=\frac{-5±\sqrt{121}}{2\times 2}$$

Tuhia te pūtakerua o te 121.

$$a=\frac{-5±11}{2\times 2}$$

Whakareatia 2 ki te 2.

$$a=\frac{-5±11}{4}$$

Nā, me whakaoti te whārite a=\frac{-5±11}{4} ina he tāpiri te ±. Tāpiri -5 ki te 11.

$$a=\frac{6}{4}$$

Whakahekea te hautanga \frac{6}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$a=\frac{3}{2}$$

Nā, me whakaoti te whārite a=\frac{-5±11}{4} ina he tango te ±. Tango 11 mai i -5.

$$a=-\frac{16}{4}$$

Whakawehe -16 ki te 4.

$$a=-4$$

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

$$2a^{2}+5a-12=2\left(a-\frac{3}{2}\right)\left(a-\left(-4\right)\right)$$

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

$$2a^{2}+5a-12=2\left(a-\frac{3}{2}\right)\left(a+4\right)$$

Tango \frac{3}{2} mai i a 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.

$$2a^{2}+5a-12=2\times \frac{2a-3}{2}\left(a+4\right)$$

Whakakorea atu te tauwehe pūnoa nui rawa 2 i roto i te 2 me te 2.

$$2a^{2}+5a-12=\left(2a-3\right)\left(a+4\right)$$

Aromātai

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