Whakaoti i te x^2-3x+2

$${ x }^{ 2 } -3x+2$$

Tauwehe

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

Ngā Upane Otinga

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

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

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 x^{2}-3x+2 hei \left(x^{2}-2x\right)+\left(-x+2\right).

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

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

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

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

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

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.

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

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.

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

Pūrua -3.

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

Whakareatia -4 ki te 2.

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

Tāpiri 9 ki te -8.

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

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-3\right)±1}{2}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±1}{2}$$

Nā, me whakaoti te whārite x=\frac{3±1}{2} ina he tāpiri te ±. Tāpiri 3 ki te 1.

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

Whakawehe 4 ki te 2.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{3±1}{2} ina he tango te ±. Tango 1 mai i 3.

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

Whakawehe 2 ki te 2.

$$x=1$$

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 2 mō te x_{1} me te 1 mō te x_{2}.

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

Aromātai

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