Whakaoti i te x^2+2x-15=

$$x ^ { 2 } + 2 x - 15 =$$

Tauwehe

$$\left(x-3\right)\left(x+5\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx-15. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=2 ab=1\left(-15\right)=-15$$

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 -15.

$$-1,15 -3,5$$

Tātaihia te tapeke mō ia takirua.

$$-1+15=14 -3+5=2$$

Ko te otinga te takirua ka hoatu i te tapeke 2.

$$a=-3 b=5$$

Tuhia anō te x^{2}+2x-15 hei \left(x^{2}-3x\right)+\left(5x-15\right).

$$\left(x^{2}-3x\right)+\left(5x-15\right)$$

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

$$x\left(x-3\right)+5\left(x-3\right)$$

Whakatauwehea atu te kīanga pātahi x-3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-3\right)\left(x+5\right)$$

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.

$$x^{2}+2x-15=0$$

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.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-15\right)}}{2}$$

Pūrua 2.

$$x=\frac{-2±\sqrt{4-4\left(-15\right)}}{2}$$

Whakareatia -4 ki te -15.

$$x=\frac{-2±\sqrt{4+60}}{2}$$

Tāpiri 4 ki te 60.

$$x=\frac{-2±\sqrt{64}}{2}$$

Tuhia te pūtakerua o te 64.

$$x=\frac{-2±8}{2}$$

Nā, me whakaoti te whārite x=\frac{-2±8}{2} ina he tāpiri te ±. Tāpiri -2 ki te 8.

$$x=\frac{6}{2}$$

Whakawehe 6 ki te 2.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{-2±8}{2} ina he tango te ±. Tango 8 mai i -2.

$$x=-\frac{10}{2}$$

Whakawehe -10 ki te 2.

$$x=-5$$

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 3 mō te x_{1} me te -5 mō te x_{2}.

$$x^{2}+2x-15=\left(x-3\right)\left(x-\left(-5\right)\right)$$

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

$$x^{2}+2x-15=\left(x-3\right)\left(x+5\right)$$

Aromātai

$$\left(x-3\right)\left(x+5\right)$$