Whakaoti i te 5x^2-41x+42

$$5 { x }^{ 2 } -41x+42$$

Tauwehe

$$\left(x-7\right)\left(5x-6\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 5x^{2}+ax+bx+42. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-41 ab=5\times 42=210$$

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 210.

$$-1,-210 -2,-105 -3,-70 -5,-42 -6,-35 -7,-30 -10,-21 -14,-15$$

Tātaihia te tapeke mō ia takirua.

$$-1-210=-211 -2-105=-107 -3-70=-73 -5-42=-47 -6-35=-41 -7-30=-37 -10-21=-31 -14-15=-29$$

Ko te otinga te takirua ka hoatu i te tapeke -41.

$$a=-35 b=-6$$

Tuhia anō te 5x^{2}-41x+42 hei \left(5x^{2}-35x\right)+\left(-6x+42\right).

$$\left(5x^{2}-35x\right)+\left(-6x+42\right)$$

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

$$5x\left(x-7\right)-6\left(x-7\right)$$

Whakatauwehea atu te kīanga pātahi x-7 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-7\right)\left(5x-6\right)$$

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.

$$5x^{2}-41x+42=0$$

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.

$$x=\frac{-\left(-41\right)±\sqrt{\left(-41\right)^{2}-4\times 5\times 42}}{2\times 5}$$

Pūrua -41.

$$x=\frac{-\left(-41\right)±\sqrt{1681-4\times 5\times 42}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-\left(-41\right)±\sqrt{1681-20\times 42}}{2\times 5}$$

Whakareatia -20 ki te 42.

$$x=\frac{-\left(-41\right)±\sqrt{1681-840}}{2\times 5}$$

Tāpiri 1681 ki te -840.

$$x=\frac{-\left(-41\right)±\sqrt{841}}{2\times 5}$$

Tuhia te pūtakerua o te 841.

$$x=\frac{-\left(-41\right)±29}{2\times 5}$$

Ko te tauaro o -41 ko 41.

$$x=\frac{41±29}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{41±29}{10}$$

Nā, me whakaoti te whārite x=\frac{41±29}{10} ina he tāpiri te ±. Tāpiri 41 ki te 29.

$$x=\frac{70}{10}$$

Whakawehe 70 ki te 10.

$$x=7$$

Nā, me whakaoti te whārite x=\frac{41±29}{10} ina he tango te ±. Tango 29 mai i 41.

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

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

$$5x^{2}-41x+42=5\left(x-7\right)\left(x-\frac{6}{5}\right)$$

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.

$$5x^{2}-41x+42=5\left(x-7\right)\times \frac{5x-6}{5}$$

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

$$5x^{2}-41x+42=\left(x-7\right)\left(5x-6\right)$$

Aromātai

$$\left(x-7\right)\left(5x-6\right)$$