Whakaoti i te x^2+6x-16

$$x ^ { 2 } + 6 x - 16$$

Tauwehe

$$\left(x-2\right)\left(x+8\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx-16. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=6 ab=1\left(-16\right)=-16$$

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 -16.

$$-1,16 -2,8 -4,4$$

Tātaihia te tapeke mō ia takirua.

$$-1+16=15 -2+8=6 -4+4=0$$

Ko te otinga te takirua ka hoatu i te tapeke 6.

$$a=-2 b=8$$

Tuhia anō te x^{2}+6x-16 hei \left(x^{2}-2x\right)+\left(8x-16\right).

$$\left(x^{2}-2x\right)+\left(8x-16\right)$$

Tauwehea te x i te tuatahi me te 8 i te rōpū tuarua.

$$x\left(x-2\right)+8\left(x-2\right)$$

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

$$\left(x-2\right)\left(x+8\right)$$

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.

$$x^{2}+6x-16=0$$

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.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-16\right)}}{2}$$

Pūrua 6.

$$x=\frac{-6±\sqrt{36-4\left(-16\right)}}{2}$$

Whakareatia -4 ki te -16.

$$x=\frac{-6±\sqrt{36+64}}{2}$$

Tāpiri 36 ki te 64.

$$x=\frac{-6±\sqrt{100}}{2}$$

Tuhia te pūtakerua o te 100.

$$x=\frac{-6±10}{2}$$

Nā, me whakaoti te whārite x=\frac{-6±10}{2} ina he tāpiri te ±. Tāpiri -6 ki te 10.

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

Whakawehe 4 ki te 2.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{-6±10}{2} ina he tango te ±. Tango 10 mai i -6.

$$x=-\frac{16}{2}$$

Whakawehe -16 ki te 2.

$$x=-8$$

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 2 mō te x_{1} me te -8 mō te x_{2}.

$$x^{2}+6x-16=\left(x-2\right)\left(x-\left(-8\right)\right)$$

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

$$x^{2}+6x-16=\left(x-2\right)\left(x+8\right)$$

Aromātai

$$\left(x-2\right)\left(x+8\right)$$