Whakaoti i te 3x^2+5x-12)-2(x^2+4x+9)

$$3 x ^ { 2 } + 5 x - 12 ) - 2 ( x ^ { 2 } + 4 x + 9 )$$

Tauwehe

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

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3x^{2}+ax+bx-12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=5 ab=3\left(-12\right)=-36$$

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 -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Tātaihia te tapeke mō ia takirua.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

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

$$a=-4 b=9$$

Tuhia anō te 3x^{2}+5x-12 hei \left(3x^{2}-4x\right)+\left(9x-12\right).

$$\left(3x^{2}-4x\right)+\left(9x-12\right)$$

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

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

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

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

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.

$$3x^{2}+5x-12=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\times 3\left(-12\right)}}{2\times 3}$$

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4\times 3\left(-12\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-5±\sqrt{25-12\left(-12\right)}}{2\times 3}$$

Whakareatia -12 ki te -12.

$$x=\frac{-5±\sqrt{25+144}}{2\times 3}$$

Tāpiri 25 ki te 144.

$$x=\frac{-5±\sqrt{169}}{2\times 3}$$

Tuhia te pūtakerua o te 169.

$$x=\frac{-5±13}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-5±13}{6}$$

Nā, me whakaoti te whārite x=\frac{-5±13}{6} ina he tāpiri te ±. Tāpiri -5 ki te 13.

$$x=\frac{8}{6}$$

Whakahekea te hautanga \frac{8}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Nā, me whakaoti te whārite x=\frac{-5±13}{6} ina he tango te ±. Tango 13 mai i -5.

$$x=-\frac{18}{6}$$

Whakawehe -18 ki te 6.

$$x=-3$$

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 -3 mō te x_{2}.

$$3x^{2}+5x-12=3\left(x-\frac{4}{3}\right)\left(x-\left(-3\right)\right)$$

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

$$3x^{2}+5x-12=3\left(x-\frac{4}{3}\right)\left(x+3\right)$$

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.

$$3x^{2}+5x-12=3\times \frac{3x-4}{3}\left(x+3\right)$$

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

$$3x^{2}+5x-12=\left(3x-4\right)\left(x+3\right)$$

Aromātai

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