Whakaoti i te f(x)=2x^2-5x+3

$$f ( x ) = 2 x ^ { 2 } - 5 x + 3$$

Tauwehe

$$\left(x-1\right)\left(2x-3\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 2x^{2}+ax+bx+3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-5 ab=2\times 3=6$$

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 6.

$$-1,-6 -2,-3$$

Tātaihia te tapeke mō ia takirua.

$$-1-6=-7 -2-3=-5$$

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

$$a=-3 b=-2$$

Tuhia anō te 2x^{2}-5x+3 hei \left(2x^{2}-3x\right)+\left(-2x+3\right).

$$\left(2x^{2}-3x\right)+\left(-2x+3\right)$$

Tauwehea te x i te tuatahi me te -1 i te rōpū tuarua.

$$x\left(2x-3\right)-\left(2x-3\right)$$

Whakatauwehea atu te kīanga pātahi 2x-3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x-3\right)\left(x-1\right)$$

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.

$$2x^{2}-5x+3=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\times 3}}{2\times 2}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\times 3}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\times 3}}{2\times 2}$$

Whakareatia -8 ki te 3.

$$x=\frac{-\left(-5\right)±\sqrt{25-24}}{2\times 2}$$

Tāpiri 25 ki te -24.

$$x=\frac{-\left(-5\right)±\sqrt{1}}{2\times 2}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-5\right)±1}{2\times 2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±1}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{5±1}{4}$$

Nā, me whakaoti te whārite x=\frac{5±1}{4} ina he tāpiri te ±. Tāpiri 5 ki te 1.

$$x=\frac{6}{4}$$

Whakahekea te hautanga \frac{6}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{3}{2}$$

Nā, me whakaoti te whārite x=\frac{5±1}{4} ina he tango te ±. Tango 1 mai i 5.

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

Whakawehe 4 ki te 4.

$$x=1$$

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{3}{2} mō te x_{1} me te 1 mō te x_{2}.

$$2x^{2}-5x+3=2\left(x-\frac{3}{2}\right)\left(x-1\right)$$

Tango \frac{3}{2} 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.

$$2x^{2}-5x+3=2\times \frac{2x-3}{2}\left(x-1\right)$$

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

$$2x^{2}-5x+3=\left(2x-3\right)\left(x-1\right)$$

Aromātai

$$\left(x-1\right)\left(2x-3\right)$$