Whakaoti i te 18x^2-15x+2

$$18 x ^ { 2 } - 15 x + 2$$

Tauwehe

$$\left(3x-2\right)\left(6x-1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 18x^{2}+ax+bx+2. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-15 ab=18\times 2=36$$

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 36.

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

Tātaihia te tapeke mō ia takirua.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

Ko te otinga te takirua ka hoatu i te tapeke -15.

$$a=-12 b=-3$$

Tuhia anō te 18x^{2}-15x+2 hei \left(18x^{2}-12x\right)+\left(-3x+2\right).

$$\left(18x^{2}-12x\right)+\left(-3x+2\right)$$

Tauwehea te 6x i te tuatahi me te -1 i te rōpū tuarua.

$$6x\left(3x-2\right)-\left(3x-2\right)$$

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

$$\left(3x-2\right)\left(6x-1\right)$$

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.

$$18x^{2}-15x+2=0$$

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.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 18\times 2}}{2\times 18}$$

Pūrua -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 18\times 2}}{2\times 18}$$

Whakareatia -4 ki te 18.

$$x=\frac{-\left(-15\right)±\sqrt{225-72\times 2}}{2\times 18}$$

Whakareatia -72 ki te 2.

$$x=\frac{-\left(-15\right)±\sqrt{225-144}}{2\times 18}$$

Tāpiri 225 ki te -144.

$$x=\frac{-\left(-15\right)±\sqrt{81}}{2\times 18}$$

Tuhia te pūtakerua o te 81.

$$x=\frac{-\left(-15\right)±9}{2\times 18}$$

Ko te tauaro o -15 ko 15.

$$x=\frac{15±9}{2\times 18}$$

Whakareatia 2 ki te 18.

$$x=\frac{15±9}{36}$$

Nā, me whakaoti te whārite x=\frac{15±9}{36} ina he tāpiri te ±. Tāpiri 15 ki te 9.

$$x=\frac{24}{36}$$

Whakahekea te hautanga \frac{24}{36} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 12.

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

Nā, me whakaoti te whārite x=\frac{15±9}{36} ina he tango te ±. Tango 9 mai i 15.

$$x=\frac{6}{36}$$

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

$$x=\frac{1}{6}$$

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

$$18x^{2}-15x+2=18\left(x-\frac{2}{3}\right)\left(x-\frac{1}{6}\right)$$

Tango \frac{2}{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.

$$18x^{2}-15x+2=18\times \frac{3x-2}{3}\left(x-\frac{1}{6}\right)$$

Tango \frac{1}{6} 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.

$$18x^{2}-15x+2=18\times \frac{3x-2}{3}\times \frac{6x-1}{6}$$

Whakareatia \frac{3x-2}{3} ki te \frac{6x-1}{6} 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.

$$18x^{2}-15x+2=18\times \frac{\left(3x-2\right)\left(6x-1\right)}{3\times 6}$$

Whakareatia 3 ki te 6.

$$18x^{2}-15x+2=18\times \frac{\left(3x-2\right)\left(6x-1\right)}{18}$$

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

$$18x^{2}-15x+2=\left(3x-2\right)\left(6x-1\right)$$

Aromātai

$$\left(3x-2\right)\left(6x-1\right)$$