Whakaoti i te 5x^2+12x+4

$$5 x ^ { 2 } + 12 x + 4$$

Tauwehe

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

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 5x^{2}+ax+bx+4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=12 ab=5\times 4=20$$

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 20.

$$1,20 2,10 4,5$$

Tātaihia te tapeke mō ia takirua.

$$1+20=21 2+10=12 4+5=9$$

Ko te otinga te takirua ka hoatu i te tapeke 12.

$$a=2 b=10$$

Tuhia anō te 5x^{2}+12x+4 hei \left(5x^{2}+2x\right)+\left(10x+4\right).

$$\left(5x^{2}+2x\right)+\left(10x+4\right)$$

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

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

Whakatauwehea atu te kīanga pātahi 5x+2 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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.

$$5x^{2}+12x+4=0$$

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.

$$x=\frac{-12±\sqrt{12^{2}-4\times 5\times 4}}{2\times 5}$$

Pūrua 12.

$$x=\frac{-12±\sqrt{144-4\times 5\times 4}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-12±\sqrt{144-20\times 4}}{2\times 5}$$

Whakareatia -20 ki te 4.

$$x=\frac{-12±\sqrt{144-80}}{2\times 5}$$

Tāpiri 144 ki te -80.

$$x=\frac{-12±\sqrt{64}}{2\times 5}$$

Tuhia te pūtakerua o te 64.

$$x=\frac{-12±8}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-12±8}{10}$$

Nā, me whakaoti te whārite x=\frac{-12±8}{10} ina he tāpiri te ±. Tāpiri -12 ki te 8.

$$x=-\frac{4}{10}$$

Whakahekea te hautanga \frac{-4}{10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Nā, me whakaoti te whārite x=\frac{-12±8}{10} ina he tango te ±. Tango 8 mai i -12.

$$x=-\frac{20}{10}$$

Whakawehe -20 ki te 10.

$$x=-2$$

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

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

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

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

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.

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

Whakakorea atu te tauwehe pūnoa nui rawa 5 i roto i te 5 me te 5.

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

Aromātai

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