Whakaoti i te 6x^2-5x-4

$$6 x ^ { 2 } - 5 x - 4$$

Tauwehe

$$\left(3x-4\right)\left(2x+1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 6x^{2}+ax+bx-4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-5 ab=6\left(-4\right)=-24$$

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 -24.

$$1,-24 2,-12 3,-8 4,-6$$

Tātaihia te tapeke mō ia takirua.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

Ko te otinga te takirua ka hoatu i te tapeke -5.

$$a=-8 b=3$$

Tuhia anō te 6x^{2}-5x-4 hei \left(6x^{2}-8x\right)+\left(3x-4\right).

$$\left(6x^{2}-8x\right)+\left(3x-4\right)$$

Whakatauwehea atu 2x i te 6x^{2}-8x.

$$2x\left(3x-4\right)+3x-4$$

Whakatauwehea atu te kīanga pātahi 3x-4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3x-4\right)\left(2x+1\right)$$

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.

$$6x^{2}-5x-4=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6\left(-4\right)}}{2\times 6}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 6\left(-4\right)}}{2\times 6}$$

Whakareatia -4 ki te 6.

$$x=\frac{-\left(-5\right)±\sqrt{25-24\left(-4\right)}}{2\times 6}$$

Whakareatia -24 ki te -4.

$$x=\frac{-\left(-5\right)±\sqrt{25+96}}{2\times 6}$$

Tāpiri 25 ki te 96.

$$x=\frac{-\left(-5\right)±\sqrt{121}}{2\times 6}$$

Tuhia te pūtakerua o te 121.

$$x=\frac{-\left(-5\right)±11}{2\times 6}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±11}{2\times 6}$$

Whakareatia 2 ki te 6.

$$x=\frac{5±11}{12}$$

Nā, me whakaoti te whārite x=\frac{5±11}{12} ina he tāpiri te ±. Tāpiri 5 ki te 11.

$$x=\frac{16}{12}$$

Whakahekea te hautanga \frac{16}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

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

Nā, me whakaoti te whārite x=\frac{5±11}{12} ina he tango te ±. Tango 11 mai i 5.

$$x=-\frac{6}{12}$$

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

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

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

$$6x^{2}-5x-4=6\left(x-\frac{4}{3}\right)\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$6x^{2}-5x-4=6\left(x-\frac{4}{3}\right)\left(x+\frac{1}{2}\right)$$

Tango \frac{4}{3} 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.

$$6x^{2}-5x-4=6\times \frac{3x-4}{3}\left(x+\frac{1}{2}\right)$$

Tāpiri \frac{1}{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.

$$6x^{2}-5x-4=6\times \frac{3x-4}{3}\times \frac{2x+1}{2}$$

Whakareatia \frac{3x-4}{3} ki te \frac{2x+1}{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.

$$6x^{2}-5x-4=6\times \frac{\left(3x-4\right)\left(2x+1\right)}{3\times 2}$$

Whakareatia 3 ki te 2.

$$6x^{2}-5x-4=6\times \frac{\left(3x-4\right)\left(2x+1\right)}{6}$$

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

$$6x^{2}-5x-4=\left(3x-4\right)\left(2x+1\right)$$

Aromātai

$$\left(3x-4\right)\left(2x+1\right)$$