Whakaoti i te 5x^2-16x+12

$$5 x ^ { 2 } - 16 x + 12$$

Tauwehe

$$\left(x-2\right)\left(5x-6\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 5x^{2}+ax+bx+12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-16 ab=5\times 12=60$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Tātaihia te tapeke mō ia takirua.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

Ko te otinga te takirua ka hoatu i te tapeke -16.

$$a=-10 b=-6$$

Tuhia anō te 5x^{2}-16x+12 hei \left(5x^{2}-10x\right)+\left(-6x+12\right).

$$\left(5x^{2}-10x\right)+\left(-6x+12\right)$$

Tauwehea te 5x i te tuatahi me te -6 i te rōpū tuarua.

$$5x\left(x-2\right)-6\left(x-2\right)$$

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

$$\left(x-2\right)\left(5x-6\right)$$

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.

$$5x^{2}-16x+12=0$$

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.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 5\times 12}}{2\times 5}$$

Pūrua -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 5\times 12}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-\left(-16\right)±\sqrt{256-20\times 12}}{2\times 5}$$

Whakareatia -20 ki te 12.

$$x=\frac{-\left(-16\right)±\sqrt{256-240}}{2\times 5}$$

Tāpiri 256 ki te -240.

$$x=\frac{-\left(-16\right)±\sqrt{16}}{2\times 5}$$

Tuhia te pūtakerua o te 16.

$$x=\frac{-\left(-16\right)±4}{2\times 5}$$

Ko te tauaro o -16 ko 16.

$$x=\frac{16±4}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{16±4}{10}$$

Nā, me whakaoti te whārite x=\frac{16±4}{10} ina he tāpiri te ±. Tāpiri 16 ki te 4.

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

Whakawehe 20 ki te 10.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{16±4}{10} ina he tango te ±. Tango 4 mai i 16.

$$x=\frac{12}{10}$$

Whakahekea te hautanga \frac{12}{10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{6}{5}$$

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 2 mō te x_{1} me te \frac{6}{5} mō te x_{2}.

$$5x^{2}-16x+12=5\left(x-2\right)\left(x-\frac{6}{5}\right)$$

Tango \frac{6}{5} 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.

$$5x^{2}-16x+12=5\left(x-2\right)\times \frac{5x-6}{5}$$

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

$$5x^{2}-16x+12=\left(x-2\right)\left(5x-6\right)$$

Aromātai

$$\left(x-2\right)\left(5x-6\right)$$