Whakaoti i te 11x^2+2x-9

$$11 x ^ { 2 } + 2 x - 9$$

Tauwehe

$$\left(11x-9\right)\left(x+1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 11x^{2}+ax+bx-9. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=2 ab=11\left(-9\right)=-99$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -99.

$$-1,99 -3,33 -9,11$$

Tātaihia te tapeke mō ia takirua.

$$-1+99=98 -3+33=30 -9+11=2$$

Ko te otinga te takirua ka hoatu i te tapeke 2.

$$a=-9 b=11$$

Tuhia anō te 11x^{2}+2x-9 hei \left(11x^{2}-9x\right)+\left(11x-9\right).

$$\left(11x^{2}-9x\right)+\left(11x-9\right)$$

Whakatauwehea atu x i te 11x^{2}-9x.

$$x\left(11x-9\right)+11x-9$$

Whakatauwehea atu te kīanga pātahi 11x-9 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(11x-9\right)\left(x+1\right)$$

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.

$$11x^{2}+2x-9=0$$

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.

$$x=\frac{-2±\sqrt{2^{2}-4\times 11\left(-9\right)}}{2\times 11}$$

Pūrua 2.

$$x=\frac{-2±\sqrt{4-4\times 11\left(-9\right)}}{2\times 11}$$

Whakareatia -4 ki te 11.

$$x=\frac{-2±\sqrt{4-44\left(-9\right)}}{2\times 11}$$

Whakareatia -44 ki te -9.

$$x=\frac{-2±\sqrt{4+396}}{2\times 11}$$

Tāpiri 4 ki te 396.

$$x=\frac{-2±\sqrt{400}}{2\times 11}$$

Tuhia te pūtakerua o te 400.

$$x=\frac{-2±20}{2\times 11}$$

Whakareatia 2 ki te 11.

$$x=\frac{-2±20}{22}$$

Nā, me whakaoti te whārite x=\frac{-2±20}{22} ina he tāpiri te ±. Tāpiri -2 ki te 20.

$$x=\frac{18}{22}$$

Whakahekea te hautanga \frac{18}{22} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{9}{11}$$

Nā, me whakaoti te whārite x=\frac{-2±20}{22} ina he tango te ±. Tango 20 mai i -2.

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

Whakawehe -22 ki te 22.

$$x=-1$$

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 \frac{9}{11} mō te x_{1} me te -1 mō te x_{2}.

$$11x^{2}+2x-9=11\left(x-\frac{9}{11}\right)\left(x-\left(-1\right)\right)$$

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

$$11x^{2}+2x-9=11\left(x-\frac{9}{11}\right)\left(x+1\right)$$

Tango \frac{9}{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.

$$11x^{2}+2x-9=11\times \frac{11x-9}{11}\left(x+1\right)$$

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

$$11x^{2}+2x-9=\left(11x-9\right)\left(x+1\right)$$

Aromātai

$$\left(11x-9\right)\left(x+1\right)$$