Whakaoti i te 12x^2-4x-1

$$12 x ^ { 2 } - 4 x - 1$$

Tauwehe

$$\left(2x-1\right)\left(6x+1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 12x^{2}+ax+bx-1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-4 ab=12\left(-1\right)=-12$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -12.

$$1,-12 2,-6 3,-4$$

Tātaihia te tapeke mō ia takirua.

$$1-12=-11 2-6=-4 3-4=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -4.

$$a=-6 b=2$$

Tuhia anō te 12x^{2}-4x-1 hei \left(12x^{2}-6x\right)+\left(2x-1\right).

$$\left(12x^{2}-6x\right)+\left(2x-1\right)$$

Whakatauwehea atu 6x i te 12x^{2}-6x.

$$6x\left(2x-1\right)+2x-1$$

Whakatauwehea atu te kīanga pātahi 2x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x-1\right)\left(6x+1\right)$$

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.

$$12x^{2}-4x-1=0$$

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.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 12\left(-1\right)}}{2\times 12}$$

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 12\left(-1\right)}}{2\times 12}$$

Whakareatia -4 ki te 12.

$$x=\frac{-\left(-4\right)±\sqrt{16-48\left(-1\right)}}{2\times 12}$$

Whakareatia -48 ki te -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+48}}{2\times 12}$$

Tāpiri 16 ki te 48.

$$x=\frac{-\left(-4\right)±\sqrt{64}}{2\times 12}$$

Tuhia te pūtakerua o te 64.

$$x=\frac{-\left(-4\right)±8}{2\times 12}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±8}{2\times 12}$$

Whakareatia 2 ki te 12.

$$x=\frac{4±8}{24}$$

Nā, me whakaoti te whārite x=\frac{4±8}{24} ina he tāpiri te ±. Tāpiri 4 ki te 8.

$$x=\frac{12}{24}$$

Whakahekea te hautanga \frac{12}{24} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 12.

$$x=\frac{1}{2}$$

Nā, me whakaoti te whārite x=\frac{4±8}{24} ina he tango te ±. Tango 8 mai i 4.

$$x=-\frac{4}{24}$$

Whakahekea te hautanga \frac{-4}{24} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x=-\frac{1}{6}$$

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}{2} mō te x_{1} me te -\frac{1}{6} mō te x_{2}.

$$12x^{2}-4x-1=12\left(x-\frac{1}{2}\right)\left(x-\left(-\frac{1}{6}\right)\right)$$

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

$$12x^{2}-4x-1=12\left(x-\frac{1}{2}\right)\left(x+\frac{1}{6}\right)$$

Tango \frac{1}{2} mai i x mā te kimi i te tauraro pātahi me te tango i ngā taurunga, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$12x^{2}-4x-1=12\times \frac{2x-1}{2}\left(x+\frac{1}{6}\right)$$

Tāpiri \frac{1}{6} 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.

$$12x^{2}-4x-1=12\times \frac{2x-1}{2}\times \frac{6x+1}{6}$$

Whakareatia \frac{2x-1}{2} ki te \frac{6x+1}{6} 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.

$$12x^{2}-4x-1=12\times \frac{\left(2x-1\right)\left(6x+1\right)}{2\times 6}$$

Whakareatia 2 ki te 6.

$$12x^{2}-4x-1=12\times \frac{\left(2x-1\right)\left(6x+1\right)}{12}$$

Whakakorea atu te tauwehe pūnoa nui rawa 12 i roto i te 12 me te 12.

$$12x^{2}-4x-1=\left(2x-1\right)\left(6x+1\right)$$

Aromātai

$$\left(2x-1\right)\left(6x+1\right)$$