Whakaoti i te 11x^2-122x+11

$$11 x ^ { 2 } - 122 x + 11$$

Tauwehe

$$\left(x-11\right)\left(11x-1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 11x^{2}+ax+bx+11. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-122 ab=11\times 11=121$$

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 121.

$$-1,-121 -11,-11$$

Tātaihia te tapeke mō ia takirua.

$$-1-121=-122 -11-11=-22$$

Ko te otinga te takirua ka hoatu i te tapeke -122.

$$a=-121 b=-1$$

Tuhia anō te 11x^{2}-122x+11 hei \left(11x^{2}-121x\right)+\left(-x+11\right).

$$\left(11x^{2}-121x\right)+\left(-x+11\right)$$

Tauwehea te 11x i te tuatahi me te -1 i te rōpū tuarua.

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

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

$$\left(x-11\right)\left(11x-1\right)$$

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.

$$11x^{2}-122x+11=0$$

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.

$$x=\frac{-\left(-122\right)±\sqrt{\left(-122\right)^{2}-4\times 11\times 11}}{2\times 11}$$

Pūrua -122.

$$x=\frac{-\left(-122\right)±\sqrt{14884-4\times 11\times 11}}{2\times 11}$$

Whakareatia -4 ki te 11.

$$x=\frac{-\left(-122\right)±\sqrt{14884-44\times 11}}{2\times 11}$$

Whakareatia -44 ki te 11.

$$x=\frac{-\left(-122\right)±\sqrt{14884-484}}{2\times 11}$$

Tāpiri 14884 ki te -484.

$$x=\frac{-\left(-122\right)±\sqrt{14400}}{2\times 11}$$

Tuhia te pūtakerua o te 14400.

$$x=\frac{-\left(-122\right)±120}{2\times 11}$$

Ko te tauaro o -122 ko 122.

$$x=\frac{122±120}{2\times 11}$$

Whakareatia 2 ki te 11.

$$x=\frac{122±120}{22}$$

Nā, me whakaoti te whārite x=\frac{122±120}{22} ina he tāpiri te ±. Tāpiri 122 ki te 120.

$$x=\frac{242}{22}$$

Whakawehe 242 ki te 22.

$$x=11$$

Nā, me whakaoti te whārite x=\frac{122±120}{22} ina he tango te ±. Tango 120 mai i 122.

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

Whakahekea te hautanga \frac{2}{22} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{1}{11}$$

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 11 mō te x_{1} me te \frac{1}{11} mō te x_{2}.

$$11x^{2}-122x+11=11\left(x-11\right)\left(x-\frac{1}{11}\right)$$

Tango \frac{1}{11} mai i x mā te kimi i te tauraro pātahi me te tango i ngā taurunga, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$11x^{2}-122x+11=11\left(x-11\right)\times \frac{11x-1}{11}$$

Whakakorea atu te tauwehe pūnoa nui rawa 11 i roto i te 11 me te 11.

$$11x^{2}-122x+11=\left(x-11\right)\left(11x-1\right)$$

Aromātai

$$\left(x-11\right)\left(11x-1\right)$$