Whakaoti i te 2d^2+9d+9

$$2 d ^ { 2 } + 9 d + 9$$

Tauwehe

$$\left(d+3\right)\left(2d+3\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 2d^{2}+ad+bd+9. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=9 ab=2\times 9=18$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 18.

$$1,18 2,9 3,6$$

Tātaihia te tapeke mō ia takirua.

$$1+18=19 2+9=11 3+6=9$$

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

$$a=3 b=6$$

Tuhia anō te 2d^{2}+9d+9 hei \left(2d^{2}+3d\right)+\left(6d+9\right).

$$\left(2d^{2}+3d\right)+\left(6d+9\right)$$

Tauwehea te d i te tuatahi me te 3 i te rōpū tuarua.

$$d\left(2d+3\right)+3\left(2d+3\right)$$

Whakatauwehea atu te kīanga pātahi 2d+3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2d+3\right)\left(d+3\right)$$

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.

$$2d^{2}+9d+9=0$$

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.

$$d=\frac{-9±\sqrt{9^{2}-4\times 2\times 9}}{2\times 2}$$

Pūrua 9.

$$d=\frac{-9±\sqrt{81-4\times 2\times 9}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$d=\frac{-9±\sqrt{81-8\times 9}}{2\times 2}$$

Whakareatia -8 ki te 9.

$$d=\frac{-9±\sqrt{81-72}}{2\times 2}$$

Tāpiri 81 ki te -72.

$$d=\frac{-9±\sqrt{9}}{2\times 2}$$

Tuhia te pūtakerua o te 9.

$$d=\frac{-9±3}{2\times 2}$$

Whakareatia 2 ki te 2.

$$d=\frac{-9±3}{4}$$

Nā, me whakaoti te whārite d=\frac{-9±3}{4} ina he tāpiri te ±. Tāpiri -9 ki te 3.

$$d=-\frac{6}{4}$$

Whakahekea te hautanga \frac{-6}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$d=-\frac{3}{2}$$

Nā, me whakaoti te whārite d=\frac{-9±3}{4} ina he tango te ±. Tango 3 mai i -9.

$$d=-\frac{12}{4}$$

Whakawehe -12 ki te 4.

$$d=-3$$

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{3}{2} mō te x_{1} me te -3 mō te x_{2}.

$$2d^{2}+9d+9=2\left(d-\left(-\frac{3}{2}\right)\right)\left(d-\left(-3\right)\right)$$

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

$$2d^{2}+9d+9=2\left(d+\frac{3}{2}\right)\left(d+3\right)$$

Tāpiri \frac{3}{2} ki te d 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.

$$2d^{2}+9d+9=2\times \frac{2d+3}{2}\left(d+3\right)$$

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

$$2d^{2}+9d+9=\left(2d+3\right)\left(d+3\right)$$

Aromātai

$$\left(d+3\right)\left(2d+3\right)$$