Whakaoti i te 9x^2+6x+1

$$9 x ^ { 2 } + 6 x + 1$$

Tauwehe

$$\left(3x+1\right)^{2}$$

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.

$$a+b=6 ab=9\times 1=9$$

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 6.

$$a=3 b=3$$

Tuhia anō te 9x^{2}+6x+1 hei \left(9x^{2}+3x\right)+\left(3x+1\right).

$$\left(9x^{2}+3x\right)+\left(3x+1\right)$$

Whakatauwehea atu 3x i te 9x^{2}+3x.

$$3x\left(3x+1\right)+3x+1$$

Whakatauwehea atu te kīanga pātahi 3x+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3x+1\right)\left(3x+1\right)$$

Tuhia anōtia hei pūrua huarua.

$$\left(3x+1\right)^{2}$$

Ngā Upane Whakamahi Tauwehe o te Pūrua Huarua

Ko te tikanga tātai o tēnei huatoru he pūrua huatoru, ka whakareatia pea e tētahi tauwehe pātahi. Ka taea ngā pūrua huatoru te tauwehe mā te kimi i ngā pūtakerua o ngā kīanga tau ārahi, autō hoki.

$$factor(9x^{2}+6x+1)$$

Kimihia te tauwehe pātahi nui rawa o ngā tau whakarea.

$$gcf(9,6,1)=1$$

Kimihia te pūtakerua o te kīanga tau ārahi, 9x^{2}.

$$\sqrt{9x^{2}}=3x$$

Ko te pūrua huatoru te pūrua o te huarua ko te tapeke tērā, te huatango rānei o ngā pūtakerua o ngā kīanga tau ārahi, autō hoki, e whakaritea ai te tohu e te tohu o te kīanga tau waenga o te pūrua huatoru.

$$\left(3x+1\right)^{2}$$

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.

$$9x^{2}+6x+1=0$$

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.

$$x=\frac{-6±\sqrt{6^{2}-4\times 9}}{2\times 9}$$

Pūrua 6.

$$x=\frac{-6±\sqrt{36-4\times 9}}{2\times 9}$$

Whakareatia -4 ki te 9.

$$x=\frac{-6±\sqrt{36-36}}{2\times 9}$$

Tāpiri 36 ki te -36.

$$x=\frac{-6±\sqrt{0}}{2\times 9}$$

Tuhia te pūtakerua o te 0.

$$x=\frac{-6±0}{2\times 9}$$

Whakareatia 2 ki te 9.

$$x=\frac{-6±0}{18}$$

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

$$9x^{2}+6x+1=9\left(x-\left(-\frac{1}{3}\right)\right)\left(x-\left(-\frac{1}{3}\right)\right)$$

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

$$9x^{2}+6x+1=9\left(x+\frac{1}{3}\right)\left(x+\frac{1}{3}\right)$$

Tāpiri \frac{1}{3} 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.

$$9x^{2}+6x+1=9\times \frac{3x+1}{3}\left(x+\frac{1}{3}\right)$$

Tāpiri \frac{1}{3} 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.

$$9x^{2}+6x+1=9\times \frac{3x+1}{3}\times \frac{3x+1}{3}$$

Whakareatia \frac{3x+1}{3} ki te \frac{3x+1}{3} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$9x^{2}+6x+1=9\times \frac{\left(3x+1\right)\left(3x+1\right)}{3\times 3}$$

Whakareatia 3 ki te 3.

$$9x^{2}+6x+1=9\times \frac{\left(3x+1\right)\left(3x+1\right)}{9}$$

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

$$9x^{2}+6x+1=\left(3x+1\right)\left(3x+1\right)$$

Aromātai

$$\left(3x+1\right)^{2}$$