Whakaoti i te 2d^2-9d-11

$$2 d ^ { 2 } - 9 d - 11$$

Tauwehe

$$\left(2d-11\right)\left(d+1\right)$$

Ngā Upane Otinga

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

$$a+b=-9 ab=2\left(-11\right)=-22$$

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 -22.

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

Tātaihia te tapeke mō ia takirua.

$$1-22=-21 2-11=-9$$

Ko te otinga te takirua ka hoatu i te tapeke -9.

$$a=-11 b=2$$

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

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

Whakatauwehea atu d i te 2d^{2}-11d.

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

Whakatauwehea atu te kīanga pātahi 2d-11 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2d-11\right)\left(d+1\right)$$

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.

$$2d^{2}-9d-11=0$$

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.

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

Pūrua -9.

$$d=\frac{-\left(-9\right)±\sqrt{81-4\times 2\left(-11\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$d=\frac{-\left(-9\right)±\sqrt{81-8\left(-11\right)}}{2\times 2}$$

Whakareatia -8 ki te -11.

$$d=\frac{-\left(-9\right)±\sqrt{81+88}}{2\times 2}$$

Tāpiri 81 ki te 88.

$$d=\frac{-\left(-9\right)±\sqrt{169}}{2\times 2}$$

Tuhia te pūtakerua o te 169.

$$d=\frac{-\left(-9\right)±13}{2\times 2}$$

Ko te tauaro o -9 ko 9.

$$d=\frac{9±13}{2\times 2}$$

Whakareatia 2 ki te 2.

$$d=\frac{9±13}{4}$$

Nā, me whakaoti te whārite d=\frac{9±13}{4} ina he tāpiri te ±. Tāpiri 9 ki te 13.

$$d=\frac{22}{4}$$

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

$$d=\frac{11}{2}$$

Nā, me whakaoti te whārite d=\frac{9±13}{4} ina he tango te ±. Tango 13 mai i 9.

$$d=-\frac{4}{4}$$

Whakawehe -4 ki te 4.

$$d=-1$$

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

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

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

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

Tango \frac{11}{2} mai i d 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.

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

Whakakorea atu te tauwehe pūnoa nui rawa 2 i roto i te 2 me te 2.

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

Aromātai

$$\left(2d-11\right)\left(d+1\right)$$