Whakaoti i te 9x^2+10x+1

$$9 { x }^{ 2 } +10x+1$$

Tauwehe

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

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 9x^{2}+ax+bx+1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=10 ab=9\times 1=9$$

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 9.

$$1,9 3,3$$

Tātaihia te tapeke mō ia takirua.

$$1+9=10 3+3=6$$

Ko te otinga te takirua ka hoatu i te tapeke 10.

$$a=1 b=9$$

Tuhia anō te 9x^{2}+10x+1 hei \left(9x^{2}+x\right)+\left(9x+1\right).

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

Whakatauwehea atu x i te 9x^{2}+x.

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

Whakatauwehea atu te kīanga pātahi 9x+1 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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.

$$9x^{2}+10x+1=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 9}}{2\times 9}$$

Pūrua 10.

$$x=\frac{-10±\sqrt{100-4\times 9}}{2\times 9}$$

Whakareatia -4 ki te 9.

$$x=\frac{-10±\sqrt{100-36}}{2\times 9}$$

Tāpiri 100 ki te -36.

$$x=\frac{-10±\sqrt{64}}{2\times 9}$$

Tuhia te pūtakerua o te 64.

$$x=\frac{-10±8}{2\times 9}$$

Whakareatia 2 ki te 9.

$$x=\frac{-10±8}{18}$$

Nā, me whakaoti te whārite x=\frac{-10±8}{18} ina he tāpiri te ±. Tāpiri -10 ki te 8.

$$x=-\frac{2}{18}$$

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

$$x=-\frac{1}{9}$$

Nā, me whakaoti te whārite x=\frac{-10±8}{18} ina he tango te ±. Tango 8 mai i -10.

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

Whakawehe -18 ki te 18.

$$x=-1$$

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{1}{9} mō te x_{1} me te -1 mō te x_{2}.

$$9x^{2}+10x+1=9\left(x-\left(-\frac{1}{9}\right)\right)\left(x-\left(-1\right)\right)$$

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

$$9x^{2}+10x+1=9\left(x+\frac{1}{9}\right)\left(x+1\right)$$

Tāpiri \frac{1}{9} 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.

$$9x^{2}+10x+1=9\times \frac{9x+1}{9}\left(x+1\right)$$

Whakakorea atu te tauwehe pūnoa nui rawa 9 i roto i te 9 me te 9.

$$9x^{2}+10x+1=\left(9x+1\right)\left(x+1\right)$$

Aromātai

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