Whakaoti i te 2j^2+11j+12

$$2 j ^ { 2 } + 11 j + 12$$

Tauwehe

$$\left(j+4\right)\left(2j+3\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 2j^{2}+aj+bj+12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=11 ab=2\times 12=24$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. 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.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

Ko te otinga te takirua ka hoatu i te tapeke 11.

$$a=3 b=8$$

Tuhia anō te 2j^{2}+11j+12 hei \left(2j^{2}+3j\right)+\left(8j+12\right).

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

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

$$j\left(2j+3\right)+4\left(2j+3\right)$$

Whakatauwehea atu te kīanga pātahi 2j+3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2j+3\right)\left(j+4\right)$$

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.

$$2j^{2}+11j+12=0$$

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.

$$j=\frac{-11±\sqrt{11^{2}-4\times 2\times 12}}{2\times 2}$$

Pūrua 11.

$$j=\frac{-11±\sqrt{121-4\times 2\times 12}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$j=\frac{-11±\sqrt{121-8\times 12}}{2\times 2}$$

Whakareatia -8 ki te 12.

$$j=\frac{-11±\sqrt{121-96}}{2\times 2}$$

Tāpiri 121 ki te -96.

$$j=\frac{-11±\sqrt{25}}{2\times 2}$$

Tuhia te pūtakerua o te 25.

$$j=\frac{-11±5}{2\times 2}$$

Whakareatia 2 ki te 2.

$$j=\frac{-11±5}{4}$$

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

$$j=-\frac{6}{4}$$

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

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

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

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

Whakawehe -16 ki te 4.

$$j=-4$$

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}.

$$2j^{2}+11j+12=2\left(j-\left(-\frac{3}{2}\right)\right)\left(j-\left(-4\right)\right)$$

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

$$2j^{2}+11j+12=2\left(j+\frac{3}{2}\right)\left(j+4\right)$$

Tāpiri \frac{3}{2} ki te j 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.

$$2j^{2}+11j+12=2\times \frac{2j+3}{2}\left(j+4\right)$$

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

$$2j^{2}+11j+12=\left(2j+3\right)\left(j+4\right)$$

Aromātai

$$\left(j+4\right)\left(2j+3\right)$$