Whakaoti i te 15x^2-4x-4=

$$15 x ^ { 2 } - 4 x - 4 =$$

Tauwehe

$$\left(3x-2\right)\left(5x+2\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 15x^{2}+ax+bx-4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-4 ab=15\left(-4\right)=-60$$

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 -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Tātaihia te tapeke mō ia takirua.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

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

$$a=-10 b=6$$

Tuhia anō te 15x^{2}-4x-4 hei \left(15x^{2}-10x\right)+\left(6x-4\right).

$$\left(15x^{2}-10x\right)+\left(6x-4\right)$$

Tauwehea te 5x i te tuatahi me te 2 i te rōpū tuarua.

$$5x\left(3x-2\right)+2\left(3x-2\right)$$

Whakatauwehea atu te kīanga pātahi 3x-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3x-2\right)\left(5x+2\right)$$

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.

$$15x^{2}-4x-4=0$$

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.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 15\left(-4\right)}}{2\times 15}$$

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 15\left(-4\right)}}{2\times 15}$$

Whakareatia -4 ki te 15.

$$x=\frac{-\left(-4\right)±\sqrt{16-60\left(-4\right)}}{2\times 15}$$

Whakareatia -60 ki te -4.

$$x=\frac{-\left(-4\right)±\sqrt{16+240}}{2\times 15}$$

Tāpiri 16 ki te 240.

$$x=\frac{-\left(-4\right)±\sqrt{256}}{2\times 15}$$

Tuhia te pūtakerua o te 256.

$$x=\frac{-\left(-4\right)±16}{2\times 15}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±16}{2\times 15}$$

Whakareatia 2 ki te 15.

$$x=\frac{4±16}{30}$$

Nā, me whakaoti te whārite x=\frac{4±16}{30} ina he tāpiri te ±. Tāpiri 4 ki te 16.

$$x=\frac{20}{30}$$

Whakahekea te hautanga \frac{20}{30} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

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

Nā, me whakaoti te whārite x=\frac{4±16}{30} ina he tango te ±. Tango 16 mai i 4.

$$x=-\frac{12}{30}$$

Whakahekea te hautanga \frac{-12}{30} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

$$x=-\frac{2}{5}$$

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{2}{3} mō te x_{1} me te -\frac{2}{5} mō te x_{2}.

$$15x^{2}-4x-4=15\left(x-\frac{2}{3}\right)\left(x-\left(-\frac{2}{5}\right)\right)$$

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

$$15x^{2}-4x-4=15\left(x-\frac{2}{3}\right)\left(x+\frac{2}{5}\right)$$

Tango \frac{2}{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.

$$15x^{2}-4x-4=15\times \frac{3x-2}{3}\left(x+\frac{2}{5}\right)$$

Tāpiri \frac{2}{5} 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.

$$15x^{2}-4x-4=15\times \frac{3x-2}{3}\times \frac{5x+2}{5}$$

Whakareatia \frac{3x-2}{3} ki te \frac{5x+2}{5} 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.

$$15x^{2}-4x-4=15\times \frac{\left(3x-2\right)\left(5x+2\right)}{3\times 5}$$

Whakareatia 3 ki te 5.

$$15x^{2}-4x-4=15\times \frac{\left(3x-2\right)\left(5x+2\right)}{15}$$

Whakakorea atu te tauwehe pūnoa nui rawa 15 i roto i te 15 me te 15.

$$15x^{2}-4x-4=\left(3x-2\right)\left(5x+2\right)$$

Aromātai

$$\left(3x-2\right)\left(5x+2\right)$$