Whakaoti i te 3x^2+5x-2

$$3 { x }^{ 2 } +5x-2$$

Tauwehe

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

Ngā Upane Otinga

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

$$a+b=5 ab=3\left(-2\right)=-6$$

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 -6.

$$-1,6 -2,3$$

Tātaihia te tapeke mō ia takirua.

$$-1+6=5 -2+3=1$$

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

$$a=-1 b=6$$

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

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

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

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

Whakatauwehea atu te kīanga pātahi 3x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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.

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

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.

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

Pūrua 5.

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

Whakareatia -4 ki te 3.

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

Whakareatia -12 ki te -2.

$$x=\frac{-5±\sqrt{25+24}}{2\times 3}$$

Tāpiri 25 ki te 24.

$$x=\frac{-5±\sqrt{49}}{2\times 3}$$

Tuhia te pūtakerua o te 49.

$$x=\frac{-5±7}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-5±7}{6}$$

Nā, me whakaoti te whārite x=\frac{-5±7}{6} ina he tāpiri te ±. Tāpiri -5 ki te 7.

$$x=\frac{2}{6}$$

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

$$x=\frac{1}{3}$$

Nā, me whakaoti te whārite x=\frac{-5±7}{6} ina he tango te ±. Tango 7 mai i -5.

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

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

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

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

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

Tango \frac{1}{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.

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

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

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

Aromātai

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