Whakaoti i te 9x^2+15x+4

$$9 x ^ { 2 } + 15 x + 4$$

Tauwehe

$$\left(3x+1\right)\left(3x+4\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 9x^{2}+ax+bx+4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=15 ab=9\times 4=36$$

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 36.

$$1,36 2,18 3,12 4,9 6,6$$

Tātaihia te tapeke mō ia takirua.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

Ko te otinga te takirua ka hoatu i te tapeke 15.

$$a=3 b=12$$

Tuhia anō te 9x^{2}+15x+4 hei \left(9x^{2}+3x\right)+\left(12x+4\right).

$$\left(9x^{2}+3x\right)+\left(12x+4\right)$$

Tauwehea te 3x i te tuatahi me te 4 i te rōpū tuarua.

$$3x\left(3x+1\right)+4\left(3x+1\right)$$

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

$$\left(3x+1\right)\left(3x+4\right)$$

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.

$$9x^{2}+15x+4=0$$

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.

$$x=\frac{-15±\sqrt{15^{2}-4\times 9\times 4}}{2\times 9}$$

Pūrua 15.

$$x=\frac{-15±\sqrt{225-4\times 9\times 4}}{2\times 9}$$

Whakareatia -4 ki te 9.

$$x=\frac{-15±\sqrt{225-36\times 4}}{2\times 9}$$

Whakareatia -36 ki te 4.

$$x=\frac{-15±\sqrt{225-144}}{2\times 9}$$

Tāpiri 225 ki te -144.

$$x=\frac{-15±\sqrt{81}}{2\times 9}$$

Tuhia te pūtakerua o te 81.

$$x=\frac{-15±9}{2\times 9}$$

Whakareatia 2 ki te 9.

$$x=\frac{-15±9}{18}$$

Nā, me whakaoti te whārite x=\frac{-15±9}{18} ina he tāpiri te ±. Tāpiri -15 ki te 9.

$$x=-\frac{6}{18}$$

Whakahekea te hautanga \frac{-6}{18} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

$$x=-\frac{1}{3}$$

Nā, me whakaoti te whārite x=\frac{-15±9}{18} ina he tango te ±. Tango 9 mai i -15.

$$x=-\frac{24}{18}$$

Whakahekea te hautanga \frac{-24}{18} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

$$x=-\frac{4}{3}$$

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{1}{3} mō te x_{1} me te -\frac{4}{3} mō te x_{2}.

$$9x^{2}+15x+4=9\left(x-\left(-\frac{1}{3}\right)\right)\left(x-\left(-\frac{4}{3}\right)\right)$$

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

$$9x^{2}+15x+4=9\left(x+\frac{1}{3}\right)\left(x+\frac{4}{3}\right)$$

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.

$$9x^{2}+15x+4=9\times \frac{3x+1}{3}\left(x+\frac{4}{3}\right)$$

Tāpiri \frac{4}{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.

$$9x^{2}+15x+4=9\times \frac{3x+1}{3}\times \frac{3x+4}{3}$$

Whakareatia \frac{3x+1}{3} ki te \frac{3x+4}{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.

$$9x^{2}+15x+4=9\times \frac{\left(3x+1\right)\left(3x+4\right)}{3\times 3}$$

Whakareatia 3 ki te 3.

$$9x^{2}+15x+4=9\times \frac{\left(3x+1\right)\left(3x+4\right)}{9}$$

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

$$9x^{2}+15x+4=\left(3x+1\right)\left(3x+4\right)$$

Aromātai

$$\left(3x+1\right)\left(3x+4\right)$$