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

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

Tauwehe

$$\left(2x-5\right)\left(x+1\right)$$

Ngā Upane Otinga

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

$$a+b=-3 ab=2\left(-5\right)=-10$$

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 -10.

$$1,-10 2,-5$$

Tātaihia te tapeke mō ia takirua.

$$1-10=-9 2-5=-3$$

Ko te otinga te takirua ka hoatu i te tapeke -3.

$$a=-5 b=2$$

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

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

Whakatauwehea atu x i te 2x^{2}-5x.

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

Whakatauwehea atu te kīanga pātahi 2x-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x-5\right)\left(x+1\right)$$

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.

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

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.

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

Pūrua -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-5\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\left(-5\right)}}{2\times 2}$$

Whakareatia -8 ki te -5.

$$x=\frac{-\left(-3\right)±\sqrt{9+40}}{2\times 2}$$

Tāpiri 9 ki te 40.

$$x=\frac{-\left(-3\right)±\sqrt{49}}{2\times 2}$$

Tuhia te pūtakerua o te 49.

$$x=\frac{-\left(-3\right)±7}{2\times 2}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±7}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{3±7}{4}$$

Nā, me whakaoti te whārite x=\frac{3±7}{4} ina he tāpiri te ±. Tāpiri 3 ki te 7.

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

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

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

Nā, me whakaoti te whārite x=\frac{3±7}{4} ina he tango te ±. Tango 7 mai i 3.

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

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{5}{2} mō te x_{1} me te -1 mō te x_{2}.

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

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

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

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.

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

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

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

Aromātai

$$\left(2x-5\right)\left(x+1\right)$$