Whakaoti i te z^2-13z+22

$$z ^ { 2 } - 13 z + 22$$

Tauwehe

$$\left(z-11\right)\left(z-2\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei z^{2}+az+bz+22. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-13 ab=1\times 22=22$$

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 22.

$$-1,-22 -2,-11$$

Tātaihia te tapeke mō ia takirua.

$$-1-22=-23 -2-11=-13$$

Ko te otinga te takirua ka hoatu i te tapeke -13.

$$a=-11 b=-2$$

Tuhia anō te z^{2}-13z+22 hei \left(z^{2}-11z\right)+\left(-2z+22\right).

$$\left(z^{2}-11z\right)+\left(-2z+22\right)$$

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

$$z\left(z-11\right)-2\left(z-11\right)$$

Whakatauwehea atu te kīanga pātahi z-11 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(z-11\right)\left(z-2\right)$$

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.

$$z^{2}-13z+22=0$$

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.

$$z=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 22}}{2}$$

Pūrua -13.

$$z=\frac{-\left(-13\right)±\sqrt{169-4\times 22}}{2}$$

Whakareatia -4 ki te 22.

$$z=\frac{-\left(-13\right)±\sqrt{169-88}}{2}$$

Tāpiri 169 ki te -88.

$$z=\frac{-\left(-13\right)±\sqrt{81}}{2}$$

Tuhia te pūtakerua o te 81.

$$z=\frac{-\left(-13\right)±9}{2}$$

Ko te tauaro o -13 ko 13.

$$z=\frac{13±9}{2}$$

Nā, me whakaoti te whārite z=\frac{13±9}{2} ina he tāpiri te ±. Tāpiri 13 ki te 9.

$$z=\frac{22}{2}$$

Whakawehe 22 ki te 2.

$$z=11$$

Nā, me whakaoti te whārite z=\frac{13±9}{2} ina he tango te ±. Tango 9 mai i 13.

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

Whakawehe 4 ki te 2.

$$z=2$$

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 11 mō te x_{1} me te 2 mō te x_{2}.

$$z^{2}-13z+22=\left(z-11\right)\left(z-2\right)$$

Aromātai

$$\left(z-11\right)\left(z-2\right)$$