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

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

Tauwehe

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

Ngā Upane Otinga

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

$$a+b=5 ab=2\times 2=4$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 4.

$$1,4 2,2$$

Tātaihia te tapeke mō ia takirua.

$$1+4=5 2+2=4$$

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

$$a=1 b=4$$

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

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

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

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

Whakatauwehea atu te kīanga pātahi 2x+1 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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.

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

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.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\times 2}}{2\times 2}$$

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4\times 2\times 2}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-5±\sqrt{25-8\times 2}}{2\times 2}$$

Whakareatia -8 ki te 2.

$$x=\frac{-5±\sqrt{25-16}}{2\times 2}$$

Tāpiri 25 ki te -16.

$$x=\frac{-5±\sqrt{9}}{2\times 2}$$

Tuhia te pūtakerua o te 9.

$$x=\frac{-5±3}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-5±3}{4}$$

Nā, me whakaoti te whārite x=\frac{-5±3}{4} ina he tāpiri te ±. Tāpiri -5 ki te 3.

$$x=-\frac{2}{4}$$

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

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

Nā, me whakaoti te whārite x=\frac{-5±3}{4} ina he tango te ±. Tango 3 mai i -5.

$$x=-\frac{8}{4}$$

Whakawehe -8 ki te 4.

$$x=-2$$

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

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

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

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

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.

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

Whakakorea atu te tauwehe pūnoa nui rawa 2 i roto i te 2 me te 2.

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

Aromātai

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