Whakaoti i te f(x)=x^2-4x+3

$$f ( x ) = x ^ { 2 } - 4 x + 3$$

Tauwehe

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

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx+3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-4 ab=1\times 3=3$$

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.

$$a=-3 b=-1$$

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

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

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

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

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

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

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.

$$x^{2}-4x+3=0$$

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.

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

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 3}}{2}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-4\right)±\sqrt{16-12}}{2}$$

Tāpiri 16 ki te -12.

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

Tuhia te pūtakerua o te 4.

$$x=\frac{-\left(-4\right)±2}{2}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±2}{2}$$

Nā, me whakaoti te whārite x=\frac{4±2}{2} ina he tāpiri te ±. Tāpiri 4 ki te 2.

$$x=\frac{6}{2}$$

Whakawehe 6 ki te 2.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{4±2}{2} ina he tango te ±. Tango 2 mai i 4.

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

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 3 mō te x_{1} me te 1 mō te x_{2}.

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

Aromātai

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