Whakaoti i te f(x)=2x^2-3x+1

$$f ( x ) = 2 x ^ { 2 } - 3 x + 1$$

Tauwehe

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

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 2x^{2}+ax+bx+1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-3 ab=2\times 1=2$$

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. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-2 b=-1$$

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

$$\left(2x^{2}-2x\right)+\left(-x+1\right)$$

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

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

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

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

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.

$$2x^{2}-3x+1=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2}}{2\times 2}$$

Pūrua -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8}}{2\times 2}$$

Tāpiri 9 ki te -8.

$$x=\frac{-\left(-3\right)±\sqrt{1}}{2\times 2}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-3\right)±1}{2\times 2}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±1}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{3±1}{4}$$

Nā, me whakaoti te whārite x=\frac{3±1}{4} ina he tāpiri te ±. Tāpiri 3 ki te 1.

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

Whakawehe 4 ki te 4.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{3±1}{4} ina he tango te ±. Tango 1 mai i 3.

$$x=\frac{2}{4}$$

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

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

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

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

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.

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

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

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

Aromātai

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