Whakaoti i te 3x^2+4x-4

$$3 x ^ { 2 } + 4 x - 4$$

Tauwehe

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

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3x^{2}+ax+bx-4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=4 ab=3\left(-4\right)=-12$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. 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.

$$a=-2 b=6$$

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

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

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

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

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

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

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.

$$3x^{2}+4x-4=0$$

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.

$$x=\frac{-4±\sqrt{4^{2}-4\times 3\left(-4\right)}}{2\times 3}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\times 3\left(-4\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-4±\sqrt{16-12\left(-4\right)}}{2\times 3}$$

Whakareatia -12 ki te -4.

$$x=\frac{-4±\sqrt{16+48}}{2\times 3}$$

Tāpiri 16 ki te 48.

$$x=\frac{-4±\sqrt{64}}{2\times 3}$$

Tuhia te pūtakerua o te 64.

$$x=\frac{-4±8}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-4±8}{6}$$

Nā, me whakaoti te whārite x=\frac{-4±8}{6} ina he tāpiri te ±. Tāpiri -4 ki te 8.

$$x=\frac{4}{6}$$

Whakahekea te hautanga \frac{4}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Nā, me whakaoti te whārite x=\frac{-4±8}{6} ina he tango te ±. Tango 8 mai i -4.

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

Whakawehe -12 ki te 6.

$$x=-2$$

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{2}{3} mō te x_{1} me te -2 mō te x_{2}.

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

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

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

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.

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

Whakakorea atu te tauwehe pūnoa nui rawa 3 i roto i te 3 me te 3.

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

Aromātai

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