Whakaoti i te h^2-8h+12

$$h ^ { 2 } - 8 h + 12$$

Tauwehe

$$\left(h-6\right)\left(h-2\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei h^{2}+ah+bh+12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-8 ab=1\times 12=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Tātaihia te tapeke mō ia takirua.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

Ko te otinga te takirua ka hoatu i te tapeke -8.

$$a=-6 b=-2$$

Tuhia anō te h^{2}-8h+12 hei \left(h^{2}-6h\right)+\left(-2h+12\right).

$$\left(h^{2}-6h\right)+\left(-2h+12\right)$$

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

$$h\left(h-6\right)-2\left(h-6\right)$$

Whakatauwehea atu te kīanga pātahi h-6 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(h-6\right)\left(h-2\right)$$

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.

$$h^{2}-8h+12=0$$

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.

$$h=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 12}}{2}$$

Pūrua -8.

$$h=\frac{-\left(-8\right)±\sqrt{64-4\times 12}}{2}$$

Whakareatia -4 ki te 12.

$$h=\frac{-\left(-8\right)±\sqrt{64-48}}{2}$$

Tāpiri 64 ki te -48.

$$h=\frac{-\left(-8\right)±\sqrt{16}}{2}$$

Tuhia te pūtakerua o te 16.

$$h=\frac{-\left(-8\right)±4}{2}$$

Ko te tauaro o -8 ko 8.

$$h=\frac{8±4}{2}$$

Nā, me whakaoti te whārite h=\frac{8±4}{2} ina he tāpiri te ±. Tāpiri 8 ki te 4.

$$h=\frac{12}{2}$$

Whakawehe 12 ki te 2.

$$h=6$$

Nā, me whakaoti te whārite h=\frac{8±4}{2} ina he tango te ±. Tango 4 mai i 8.

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

Whakawehe 4 ki te 2.

$$h=2$$

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 6 mō te x_{1} me te 2 mō te x_{2}.

$$h^{2}-8h+12=\left(h-6\right)\left(h-2\right)$$

Aromātai

$$\left(h-6\right)\left(h-2\right)$$