Whakaoti i te 6a^2-5a+1

$$6 a ^ { 2 } - 5 a + 1$$

Tauwehe

$$\left(2a-1\right)\left(3a-1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 6a^{2}+pa+qa+1. Hei kimi p me q, whakaritea tētahi pūnaha kia whakaoti.

$$p+q=-5 pq=6\times 1=6$$

I te mea kua tōrunga te pq, he ōrite te tohu o p me q. I te mea kua tōraro te p+q, he tōraro hoki a p me q. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 6.

$$-1,-6 -2,-3$$

Tātaihia te tapeke mō ia takirua.

$$-1-6=-7 -2-3=-5$$

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

$$p=-3 q=-2$$

Tuhia anō te 6a^{2}-5a+1 hei \left(6a^{2}-3a\right)+\left(-2a+1\right).

$$\left(6a^{2}-3a\right)+\left(-2a+1\right)$$

Tauwehea te 3a i te tuatahi me te -1 i te rōpū tuarua.

$$3a\left(2a-1\right)-\left(2a-1\right)$$

Whakatauwehea atu te kīanga pātahi 2a-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2a-1\right)\left(3a-1\right)$$

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.

$$6a^{2}-5a+1=0$$

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.

$$a=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6}}{2\times 6}$$

Pūrua -5.

$$a=\frac{-\left(-5\right)±\sqrt{25-4\times 6}}{2\times 6}$$

Whakareatia -4 ki te 6.

$$a=\frac{-\left(-5\right)±\sqrt{25-24}}{2\times 6}$$

Tāpiri 25 ki te -24.

$$a=\frac{-\left(-5\right)±\sqrt{1}}{2\times 6}$$

Tuhia te pūtakerua o te 1.

$$a=\frac{-\left(-5\right)±1}{2\times 6}$$

Ko te tauaro o -5 ko 5.

$$a=\frac{5±1}{2\times 6}$$

Whakareatia 2 ki te 6.

$$a=\frac{5±1}{12}$$

Nā, me whakaoti te whārite a=\frac{5±1}{12} ina he tāpiri te ±. Tāpiri 5 ki te 1.

$$a=\frac{6}{12}$$

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

$$a=\frac{1}{2}$$

Nā, me whakaoti te whārite a=\frac{5±1}{12} ina he tango te ±. Tango 1 mai i 5.

$$a=\frac{4}{12}$$

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

$$a=\frac{1}{3}$$

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

$$6a^{2}-5a+1=6\left(a-\frac{1}{2}\right)\left(a-\frac{1}{3}\right)$$

Tango \frac{1}{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.

$$6a^{2}-5a+1=6\times \frac{2a-1}{2}\left(a-\frac{1}{3}\right)$$

Tango \frac{1}{3} 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.

$$6a^{2}-5a+1=6\times \frac{2a-1}{2}\times \frac{3a-1}{3}$$

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

$$6a^{2}-5a+1=6\times \frac{\left(2a-1\right)\left(3a-1\right)}{2\times 3}$$

Whakareatia 2 ki te 3.

$$6a^{2}-5a+1=6\times \frac{\left(2a-1\right)\left(3a-1\right)}{6}$$

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

$$6a^{2}-5a+1=\left(2a-1\right)\left(3a-1\right)$$

Aromātai

$$\left(2a-1\right)\left(3a-1\right)$$