Whakaoti i te 2x^2-13x+20=

$$2 x ^ { 2 } - 13 x + 20 =$$

Tauwehe

$$\left(x-4\right)\left(2x-5\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 2x^{2}+ax+bx+20. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-13 ab=2\times 20=40$$

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 40.

$$-1,-40 -2,-20 -4,-10 -5,-8$$

Tātaihia te tapeke mō ia takirua.

$$-1-40=-41 -2-20=-22 -4-10=-14 -5-8=-13$$

Ko te otinga te takirua ka hoatu i te tapeke -13.

$$a=-8 b=-5$$

Tuhia anō te 2x^{2}-13x+20 hei \left(2x^{2}-8x\right)+\left(-5x+20\right).

$$\left(2x^{2}-8x\right)+\left(-5x+20\right)$$

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

$$2x\left(x-4\right)-5\left(x-4\right)$$

Whakatauwehea atu te kīanga pātahi x-4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-4\right)\left(2x-5\right)$$

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.

$$2x^{2}-13x+20=0$$

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.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 2\times 20}}{2\times 2}$$

Pūrua -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 2\times 20}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-13\right)±\sqrt{169-8\times 20}}{2\times 2}$$

Whakareatia -8 ki te 20.

$$x=\frac{-\left(-13\right)±\sqrt{169-160}}{2\times 2}$$

Tāpiri 169 ki te -160.

$$x=\frac{-\left(-13\right)±\sqrt{9}}{2\times 2}$$

Tuhia te pūtakerua o te 9.

$$x=\frac{-\left(-13\right)±3}{2\times 2}$$

Ko te tauaro o -13 ko 13.

$$x=\frac{13±3}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{13±3}{4}$$

Nā, me whakaoti te whārite x=\frac{13±3}{4} ina he tāpiri te ±. Tāpiri 13 ki te 3.

$$x=\frac{16}{4}$$

Whakawehe 16 ki te 4.

$$x=4$$

Nā, me whakaoti te whārite x=\frac{13±3}{4} ina he tango te ±. Tango 3 mai i 13.

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

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

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

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

$$2x^{2}-13x+20=2\left(x-4\right)\left(x-\frac{5}{2}\right)$$

Tango \frac{5}{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.

$$2x^{2}-13x+20=2\left(x-4\right)\times \frac{2x-5}{2}$$

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

$$2x^{2}-13x+20=\left(x-4\right)\left(2x-5\right)$$

Aromātai

$$\left(x-4\right)\left(2x-5\right)$$