Whakaoti i te 2x^2+9x-5

$$2 { x }^{ 2 } +9x-5$$

Tauwehe

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

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.

$$a+b=9 ab=2\left(-5\right)=-10$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. 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 9.

$$a=-1 b=10$$

Tuhia anō te 2x^{2}+9x-5 hei \left(2x^{2}-x\right)+\left(10x-5\right).

$$\left(2x^{2}-x\right)+\left(10x-5\right)$$

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

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

Whakatauwehea atu te kīanga pātahi 2x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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.

$$2x^{2}+9x-5=0$$

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.

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

Pūrua 9.

$$x=\frac{-9±\sqrt{81-4\times 2\left(-5\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-9±\sqrt{81-8\left(-5\right)}}{2\times 2}$$

Whakareatia -8 ki te -5.

$$x=\frac{-9±\sqrt{81+40}}{2\times 2}$$

Tāpiri 81 ki te 40.

$$x=\frac{-9±\sqrt{121}}{2\times 2}$$

Tuhia te pūtakerua o te 121.

$$x=\frac{-9±11}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-9±11}{4}$$

Nā, me whakaoti te whārite x=\frac{-9±11}{4} ina he tāpiri te ±. Tāpiri -9 ki te 11.

$$x=\frac{2}{4}$$

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

$$x=\frac{1}{2}$$

Nā, me whakaoti te whārite x=\frac{-9±11}{4} ina he tango te ±. Tango 11 mai i -9.

$$x=-\frac{20}{4}$$

Whakawehe -20 ki te 4.

$$x=-5$$

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

$$2x^{2}+9x-5=2\left(x-\frac{1}{2}\right)\left(x-\left(-5\right)\right)$$

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

$$2x^{2}+9x-5=2\left(x-\frac{1}{2}\right)\left(x+5\right)$$

Tango \frac{1}{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.

$$2x^{2}+9x-5=2\times \frac{2x-1}{2}\left(x+5\right)$$

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

$$2x^{2}+9x-5=\left(2x-1\right)\left(x+5\right)$$

Aromātai

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