Whakaoti i te 6x^2-17x+12

$$6 x ^ { 2 } - 17 x + 12$$

Tauwehe

$$\left(2x-3\right)\left(3x-4\right)$$

Ngā Upane Otinga

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

$$a+b=-17 ab=6\times 12=72$$

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

$$-1-72=-73 -2-36=-38 -3-24=-27 -4-18=-22 -6-12=-18 -8-9=-17$$

Ko te otinga te takirua ka hoatu i te tapeke -17.

$$a=-9 b=-8$$

Tuhia anō te 6x^{2}-17x+12 hei \left(6x^{2}-9x\right)+\left(-8x+12\right).

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

Tauwehea te 3x i te tuatahi me te -4 i te rōpū tuarua.

$$3x\left(2x-3\right)-4\left(2x-3\right)$$

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

$$\left(2x-3\right)\left(3x-4\right)$$

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.

$$6x^{2}-17x+12=0$$

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.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 6\times 12}}{2\times 6}$$

Pūrua -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 6\times 12}}{2\times 6}$$

Whakareatia -4 ki te 6.

$$x=\frac{-\left(-17\right)±\sqrt{289-24\times 12}}{2\times 6}$$

Whakareatia -24 ki te 12.

$$x=\frac{-\left(-17\right)±\sqrt{289-288}}{2\times 6}$$

Tāpiri 289 ki te -288.

$$x=\frac{-\left(-17\right)±\sqrt{1}}{2\times 6}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-17\right)±1}{2\times 6}$$

Ko te tauaro o -17 ko 17.

$$x=\frac{17±1}{2\times 6}$$

Whakareatia 2 ki te 6.

$$x=\frac{17±1}{12}$$

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

$$x=\frac{18}{12}$$

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

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

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

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

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

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

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

$$6x^{2}-17x+12=6\left(x-\frac{3}{2}\right)\left(x-\frac{4}{3}\right)$$

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

$$6x^{2}-17x+12=6\times \frac{2x-3}{2}\left(x-\frac{4}{3}\right)$$

Tango \frac{4}{3} mai i x 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.

$$6x^{2}-17x+12=6\times \frac{2x-3}{2}\times \frac{3x-4}{3}$$

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

$$6x^{2}-17x+12=6\times \frac{\left(2x-3\right)\left(3x-4\right)}{2\times 3}$$

Whakareatia 2 ki te 3.

$$6x^{2}-17x+12=6\times \frac{\left(2x-3\right)\left(3x-4\right)}{6}$$

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

$$6x^{2}-17x+12=\left(2x-3\right)\left(3x-4\right)$$

Aromātai

$$\left(2x-3\right)\left(3x-4\right)$$