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 tōraro te a+b, he tōraro 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.

$$a=-10 b=-2$$

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

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

Tauwehea te 5x i te tuatahi me te -2 i te rōpū tuarua.

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

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

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

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.

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

Pūrua -12.

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

Whakareatia -4 ki te 5.

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

Whakareatia -20 ki te 4.

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

Tāpiri 144 ki te -80.

$$x=\frac{-\left(-12\right)±\sqrt{64}}{2\times 5}$$

Tuhia te pūtakerua o te 64.

$$x=\frac{-\left(-12\right)±8}{2\times 5}$$

Ko te tauaro o -12 ko 12.

$$x=\frac{12±8}{2\times 5}$$

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.

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

Whakawehe 20 ki te 10.

$$x=2$$

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

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

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

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

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{2}{5} mō te x_{2}.

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

Tango \frac{2}{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.

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

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

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

Aromātai

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