Whakaoti i te 6x^2+11x+3=

$$6 x ^ { 2 } + 11 x + 3 =$$

Tauwehe

$$\left(2x+3\right)\left(3x+1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 6x^{2}+ax+bx+3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=11 ab=6\times 3=18$$

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 11.

$$a=2 b=9$$

Tuhia anō te 6x^{2}+11x+3 hei \left(6x^{2}+2x\right)+\left(9x+3\right).

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

Tauwehea te 2x i te tuatahi me te 3 i te rōpū tuarua.

$$2x\left(3x+1\right)+3\left(3x+1\right)$$

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

$$\left(3x+1\right)\left(2x+3\right)$$

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.

$$6x^{2}+11x+3=0$$

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.

$$x=\frac{-11±\sqrt{11^{2}-4\times 6\times 3}}{2\times 6}$$

Pūrua 11.

$$x=\frac{-11±\sqrt{121-4\times 6\times 3}}{2\times 6}$$

Whakareatia -4 ki te 6.

$$x=\frac{-11±\sqrt{121-24\times 3}}{2\times 6}$$

Whakareatia -24 ki te 3.

$$x=\frac{-11±\sqrt{121-72}}{2\times 6}$$

Tāpiri 121 ki te -72.

$$x=\frac{-11±\sqrt{49}}{2\times 6}$$

Tuhia te pūtakerua o te 49.

$$x=\frac{-11±7}{2\times 6}$$

Whakareatia 2 ki te 6.

$$x=\frac{-11±7}{12}$$

Nā, me whakaoti te whārite x=\frac{-11±7}{12} ina he tāpiri te ±. Tāpiri -11 ki te 7.

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

Whakahekea te hautanga \frac{-4}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

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

Nā, me whakaoti te whārite x=\frac{-11±7}{12} ina he tango te ±. Tango 7 mai i -11.

$$x=-\frac{18}{12}$$

Whakahekea te hautanga \frac{-18}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

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

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{3}{2} mō te x_{2}.

$$6x^{2}+11x+3=6\left(x-\left(-\frac{1}{3}\right)\right)\left(x-\left(-\frac{3}{2}\right)\right)$$

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

$$6x^{2}+11x+3=6\left(x+\frac{1}{3}\right)\left(x+\frac{3}{2}\right)$$

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.

$$6x^{2}+11x+3=6\times \frac{3x+1}{3}\left(x+\frac{3}{2}\right)$$

Tāpiri \frac{3}{2} 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.

$$6x^{2}+11x+3=6\times \frac{3x+1}{3}\times \frac{2x+3}{2}$$

Whakareatia \frac{3x+1}{3} ki te \frac{2x+3}{2} 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.

$$6x^{2}+11x+3=6\times \frac{\left(3x+1\right)\left(2x+3\right)}{3\times 2}$$

Whakareatia 3 ki te 2.

$$6x^{2}+11x+3=6\times \frac{\left(3x+1\right)\left(2x+3\right)}{6}$$

Whakakorea atu te tauwehe pūnoa nui rawa 6 i roto i te 6 me te 6.

$$6x^{2}+11x+3=\left(3x+1\right)\left(2x+3\right)$$

Aromātai

$$\left(2x+3\right)\left(3x+1\right)$$