Whakaoti i te 6-5n+n^2

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

Tauwehe

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

Ngā Upane Otinga

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

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

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei n^{2}+an+bn+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 n^{2}-5n+6 hei \left(n^{2}-3n\right)+\left(-2n+6\right).

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

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

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

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

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

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.

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

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.

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

Pūrua -5.

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

Whakareatia -4 ki te 6.

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

Tāpiri 25 ki te -24.

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

Tuhia te pūtakerua o te 1.

$$n=\frac{-\left(-5\right)±1}{2}$$

Ko te tauaro o -5 ko 5.

$$n=\frac{5±1}{2}$$

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

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

Whakawehe 6 ki te 2.

$$n=3$$

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

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

Whakawehe 4 ki te 2.

$$n=2$$

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}.

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

Aromātai

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