Whakaoti i te 6x^2-11x+4

$$6 { x }^{ 2 } -11x+4$$

Tauwehe

$$\left(2x-1\right)\left(3x-4\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 6x^{2}+ax+bx+4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-11 ab=6\times 4=24$$

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 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.

$$a=-8 b=-3$$

Tuhia anō te 6x^{2}-11x+4 hei \left(6x^{2}-8x\right)+\left(-3x+4\right).

$$\left(6x^{2}-8x\right)+\left(-3x+4\right)$$

Tauwehea te 2x i te tuatahi me te -1 i te rōpū tuarua.

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

Whakatauwehea atu te kīanga pātahi 3x-4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3x-4\right)\left(2x-1\right)$$

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.

$$6x^{2}-11x+4=0$$

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.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 6\times 4}}{2\times 6}$$

Pūrua -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 6\times 4}}{2\times 6}$$

Whakareatia -4 ki te 6.

$$x=\frac{-\left(-11\right)±\sqrt{121-24\times 4}}{2\times 6}$$

Whakareatia -24 ki te 4.

$$x=\frac{-\left(-11\right)±\sqrt{121-96}}{2\times 6}$$

Tāpiri 121 ki te -96.

$$x=\frac{-\left(-11\right)±\sqrt{25}}{2\times 6}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-\left(-11\right)±5}{2\times 6}$$

Ko te tauaro o -11 ko 11.

$$x=\frac{11±5}{2\times 6}$$

Whakareatia 2 ki te 6.

$$x=\frac{11±5}{12}$$

Nā, me whakaoti te whārite x=\frac{11±5}{12} ina he tāpiri te ±. Tāpiri 11 ki te 5.

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

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

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

Nā, me whakaoti te whārite x=\frac{11±5}{12} ina he tango te ±. Tango 5 mai i 11.

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

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

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

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

$$6x^{2}-11x+4=6\left(x-\frac{4}{3}\right)\left(x-\frac{1}{2}\right)$$

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.

$$6x^{2}-11x+4=6\times \frac{3x-4}{3}\left(x-\frac{1}{2}\right)$$

Tango \frac{1}{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.

$$6x^{2}-11x+4=6\times \frac{3x-4}{3}\times \frac{2x-1}{2}$$

Whakareatia \frac{3x-4}{3} ki te \frac{2x-1}{2} 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.

$$6x^{2}-11x+4=6\times \frac{\left(3x-4\right)\left(2x-1\right)}{3\times 2}$$

Whakareatia 3 ki te 2.

$$6x^{2}-11x+4=6\times \frac{\left(3x-4\right)\left(2x-1\right)}{6}$$

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

$$6x^{2}-11x+4=\left(3x-4\right)\left(2x-1\right)$$

Aromātai

$$\left(2x-1\right)\left(3x-4\right)$$