Whakaoti i te 2y^2-5y+2

$$2 y ^ { 2 } - 5 y + 2$$

Tauwehe

$$\left(y-2\right)\left(2y-1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 2y^{2}+ay+by+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 tōraro te a+b, he tōraro 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.

$$a=-4 b=-1$$

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

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

Tauwehea te 2y i te tuatahi me te -1 i te rōpū tuarua.

$$2y\left(y-2\right)-\left(y-2\right)$$

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

$$\left(y-2\right)\left(2y-1\right)$$

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.

$$2y^{2}-5y+2=0$$

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.

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

Pūrua -5.

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

Whakareatia -4 ki te 2.

$$y=\frac{-\left(-5\right)±\sqrt{25-8\times 2}}{2\times 2}$$

Whakareatia -8 ki te 2.

$$y=\frac{-\left(-5\right)±\sqrt{25-16}}{2\times 2}$$

Tāpiri 25 ki te -16.

$$y=\frac{-\left(-5\right)±\sqrt{9}}{2\times 2}$$

Tuhia te pūtakerua o te 9.

$$y=\frac{-\left(-5\right)±3}{2\times 2}$$

Ko te tauaro o -5 ko 5.

$$y=\frac{5±3}{2\times 2}$$

Whakareatia 2 ki te 2.

$$y=\frac{5±3}{4}$$

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

$$y=\frac{8}{4}$$

Whakawehe 8 ki te 4.

$$y=2$$

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

$$y=\frac{2}{4}$$

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

$$y=\frac{1}{2}$$

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

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

Tango \frac{1}{2} mai i y 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.

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

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

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

Aromātai

$$\left(y-2\right)\left(2y-1\right)$$