Whakaoti i te 11x^2-20x-4

$$11 x ^ { 2 } - 20 x - 4$$

Tauwehe

$$\left(x-2\right)\left(11x+2\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 11x^{2}+ax+bx-4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-20 ab=11\left(-4\right)=-44$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -44.

$$1,-44 2,-22 4,-11$$

Tātaihia te tapeke mō ia takirua.

$$1-44=-43 2-22=-20 4-11=-7$$

Ko te otinga te takirua ka hoatu i te tapeke -20.

$$a=-22 b=2$$

Tuhia anō te 11x^{2}-20x-4 hei \left(11x^{2}-22x\right)+\left(2x-4\right).

$$\left(11x^{2}-22x\right)+\left(2x-4\right)$$

Tauwehea te 11x i te tuatahi me te 2 i te rōpū tuarua.

$$11x\left(x-2\right)+2\left(x-2\right)$$

Whakatauwehea atu te kīanga pātahi x-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-2\right)\left(11x+2\right)$$

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.

$$11x^{2}-20x-4=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 11\left(-4\right)}}{2\times 11}$$

Pūrua -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 11\left(-4\right)}}{2\times 11}$$

Whakareatia -4 ki te 11.

$$x=\frac{-\left(-20\right)±\sqrt{400-44\left(-4\right)}}{2\times 11}$$

Whakareatia -44 ki te -4.

$$x=\frac{-\left(-20\right)±\sqrt{400+176}}{2\times 11}$$

Tāpiri 400 ki te 176.

$$x=\frac{-\left(-20\right)±\sqrt{576}}{2\times 11}$$

Tuhia te pūtakerua o te 576.

$$x=\frac{-\left(-20\right)±24}{2\times 11}$$

Ko te tauaro o -20 ko 20.

$$x=\frac{20±24}{2\times 11}$$

Whakareatia 2 ki te 11.

$$x=\frac{20±24}{22}$$

Nā, me whakaoti te whārite x=\frac{20±24}{22} ina he tāpiri te ±. Tāpiri 20 ki te 24.

$$x=\frac{44}{22}$$

Whakawehe 44 ki te 22.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{20±24}{22} ina he tango te ±. Tango 24 mai i 20.

$$x=-\frac{4}{22}$$

Whakahekea te hautanga \frac{-4}{22} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{2}{11}$$

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

$$11x^{2}-20x-4=11\left(x-2\right)\left(x-\left(-\frac{2}{11}\right)\right)$$

Whakamāmātia ngā kīanga katoa o te āhua p-\left(-q\right) ki te p+q.

$$11x^{2}-20x-4=11\left(x-2\right)\left(x+\frac{2}{11}\right)$$

Tāpiri \frac{2}{11} ki te x mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$11x^{2}-20x-4=11\left(x-2\right)\times \frac{11x+2}{11}$$

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

$$11x^{2}-20x-4=\left(x-2\right)\left(11x+2\right)$$

Aromātai

$$\left(x-2\right)\left(11x+2\right)$$