Whakaoti i te x^2-5x+6

$$x ^ { 2 } -5x+6$$

Tauwehe

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

Ngā Upane Otinga

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

$$a+b=-5 ab=1\times 6=6$$

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 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.

$$a=-3 b=-2$$

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

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

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

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

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

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

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.

$$x^{2}-5x+6=0$$

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.

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

Pūrua -5.

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

Whakareatia -4 ki te 6.

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

Tāpiri 25 ki te -24.

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

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-5\right)±1}{2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±1}{2}$$

Nā, me whakaoti te whārite x=\frac{5±1}{2} ina he tāpiri te ±. Tāpiri 5 ki te 1.

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

Whakawehe 6 ki te 2.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{5±1}{2} ina he tango te ±. Tango 1 mai i 5.

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

Whakawehe 4 ki te 2.

$$x=2$$

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 3 mō te x_{1} me te 2 mō te x_{2}.

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

Aromātai

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