Whakaoti i te q^2-10q+21

$$q ^ { 2 } - 10 q + 21$$

Tauwehe

$$\left(q-7\right)\left(q-3\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei q^{2}+aq+bq+21. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-10 ab=1\times 21=21$$

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 21.

$$-1,-21 -3,-7$$

Tātaihia te tapeke mō ia takirua.

$$-1-21=-22 -3-7=-10$$

Ko te otinga te takirua ka hoatu i te tapeke -10.

$$a=-7 b=-3$$

Tuhia anō te q^{2}-10q+21 hei \left(q^{2}-7q\right)+\left(-3q+21\right).

$$\left(q^{2}-7q\right)+\left(-3q+21\right)$$

Tauwehea te q i te tuatahi me te -3 i te rōpū tuarua.

$$q\left(q-7\right)-3\left(q-7\right)$$

Whakatauwehea atu te kīanga pātahi q-7 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(q-7\right)\left(q-3\right)$$

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.

$$q^{2}-10q+21=0$$

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.

$$q=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 21}}{2}$$

Pūrua -10.

$$q=\frac{-\left(-10\right)±\sqrt{100-4\times 21}}{2}$$

Whakareatia -4 ki te 21.

$$q=\frac{-\left(-10\right)±\sqrt{100-84}}{2}$$

Tāpiri 100 ki te -84.

$$q=\frac{-\left(-10\right)±\sqrt{16}}{2}$$

Tuhia te pūtakerua o te 16.

$$q=\frac{-\left(-10\right)±4}{2}$$

Ko te tauaro o -10 ko 10.

$$q=\frac{10±4}{2}$$

Nā, me whakaoti te whārite q=\frac{10±4}{2} ina he tāpiri te ±. Tāpiri 10 ki te 4.

$$q=\frac{14}{2}$$

Whakawehe 14 ki te 2.

$$q=7$$

Nā, me whakaoti te whārite q=\frac{10±4}{2} ina he tango te ±. Tango 4 mai i 10.

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

Whakawehe 6 ki te 2.

$$q=3$$

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 7 mō te x_{1} me te 3 mō te x_{2}.

$$q^{2}-10q+21=\left(q-7\right)\left(q-3\right)$$

Aromātai

$$\left(q-7\right)\left(q-3\right)$$