Whakaoti i te 7x^2-5x-2

$$7 x ^ { 2 } - 5 x - 2$$

Tauwehe

$$\left(x-1\right)\left(7x+2\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 7x^{2}+ax+bx-2. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-5 ab=7\left(-2\right)=-14$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -14.

$$1,-14 2,-7$$

Tātaihia te tapeke mō ia takirua.

$$1-14=-13 2-7=-5$$

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

$$a=-7 b=2$$

Tuhia anō te 7x^{2}-5x-2 hei \left(7x^{2}-7x\right)+\left(2x-2\right).

$$\left(7x^{2}-7x\right)+\left(2x-2\right)$$

Tauwehea te 7x i te tuatahi me te 2 i te rōpū tuarua.

$$7x\left(x-1\right)+2\left(x-1\right)$$

Whakatauwehea atu te kīanga pātahi x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-1\right)\left(7x+2\right)$$

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.

$$7x^{2}-5x-2=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 7\left(-2\right)}}{2\times 7}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 7\left(-2\right)}}{2\times 7}$$

Whakareatia -4 ki te 7.

$$x=\frac{-\left(-5\right)±\sqrt{25-28\left(-2\right)}}{2\times 7}$$

Whakareatia -28 ki te -2.

$$x=\frac{-\left(-5\right)±\sqrt{25+56}}{2\times 7}$$

Tāpiri 25 ki te 56.

$$x=\frac{-\left(-5\right)±\sqrt{81}}{2\times 7}$$

Tuhia te pūtakerua o te 81.

$$x=\frac{-\left(-5\right)±9}{2\times 7}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±9}{2\times 7}$$

Whakareatia 2 ki te 7.

$$x=\frac{5±9}{14}$$

Nā, me whakaoti te whārite x=\frac{5±9}{14} ina he tāpiri te ±. Tāpiri 5 ki te 9.

$$x=\frac{14}{14}$$

Whakawehe 14 ki te 14.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{5±9}{14} ina he tango te ±. Tango 9 mai i 5.

$$x=-\frac{4}{14}$$

Whakahekea te hautanga \frac{-4}{14} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{2}{7}$$

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

$$7x^{2}-5x-2=7\left(x-1\right)\left(x-\left(-\frac{2}{7}\right)\right)$$

Whakamāmātia ngā kīanga katoa o te āhua p-\left(-q\right) ki te p+q.

$$7x^{2}-5x-2=7\left(x-1\right)\left(x+\frac{2}{7}\right)$$

Tāpiri \frac{2}{7} ki te x mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$7x^{2}-5x-2=7\left(x-1\right)\times \frac{7x+2}{7}$$

Whakakorea atu te tauwehe pūnoa nui rawa 7 i roto i te 7 me te 7.

$$7x^{2}-5x-2=\left(x-1\right)\left(7x+2\right)$$

Aromātai

$$\left(x-1\right)\left(7x+2\right)$$