Whakaoti i te x^2-23x+132

$$x ^ { 2 } - 23 x + 132$$

Tauwehe

$$\left(x-12\right)\left(x-11\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx+132. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-23 ab=1\times 132=132$$

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 132.

$$-1,-132 -2,-66 -3,-44 -4,-33 -6,-22 -11,-12$$

Tātaihia te tapeke mō ia takirua.

$$-1-132=-133 -2-66=-68 -3-44=-47 -4-33=-37 -6-22=-28 -11-12=-23$$

Ko te otinga te takirua ka hoatu i te tapeke -23.

$$a=-12 b=-11$$

Tuhia anō te x^{2}-23x+132 hei \left(x^{2}-12x\right)+\left(-11x+132\right).

$$\left(x^{2}-12x\right)+\left(-11x+132\right)$$

Tauwehea te x i te tuatahi me te -11 i te rōpū tuarua.

$$x\left(x-12\right)-11\left(x-12\right)$$

Whakatauwehea atu te kīanga pātahi x-12 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-12\right)\left(x-11\right)$$

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.

$$x^{2}-23x+132=0$$

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.

$$x=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 132}}{2}$$

Pūrua -23.

$$x=\frac{-\left(-23\right)±\sqrt{529-4\times 132}}{2}$$

Whakareatia -4 ki te 132.

$$x=\frac{-\left(-23\right)±\sqrt{529-528}}{2}$$

Tāpiri 529 ki te -528.

$$x=\frac{-\left(-23\right)±\sqrt{1}}{2}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-23\right)±1}{2}$$

Ko te tauaro o -23 ko 23.

$$x=\frac{23±1}{2}$$

Nā, me whakaoti te whārite x=\frac{23±1}{2} ina he tāpiri te ±. Tāpiri 23 ki te 1.

$$x=\frac{24}{2}$$

Whakawehe 24 ki te 2.

$$x=12$$

Nā, me whakaoti te whārite x=\frac{23±1}{2} ina he tango te ±. Tango 1 mai i 23.

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

Whakawehe 22 ki te 2.

$$x=11$$

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 12 mō te x_{1} me te 11 mō te x_{2}.

$$x^{2}-23x+132=\left(x-12\right)\left(x-11\right)$$

Aromātai

$$\left(x-12\right)\left(x-11\right)$$