Whakaoti i te x^2+4x-117

$${ x }^{ 2 } +4x-117$$

Tauwehe

$$\left(x-9\right)\left(x+13\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx-117. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=4 ab=1\left(-117\right)=-117$$

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 -117.

$$-1,117 -3,39 -9,13$$

Tātaihia te tapeke mō ia takirua.

$$-1+117=116 -3+39=36 -9+13=4$$

Ko te otinga te takirua ka hoatu i te tapeke 4.

$$a=-9 b=13$$

Tuhia anō te x^{2}+4x-117 hei \left(x^{2}-9x\right)+\left(13x-117\right).

$$\left(x^{2}-9x\right)+\left(13x-117\right)$$

Tauwehea te x i te tuatahi me te 13 i te rōpū tuarua.

$$x\left(x-9\right)+13\left(x-9\right)$$

Whakatauwehea atu te kīanga pātahi x-9 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-9\right)\left(x+13\right)$$

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.

$$x^{2}+4x-117=0$$

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.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-117\right)}}{2}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\left(-117\right)}}{2}$$

Whakareatia -4 ki te -117.

$$x=\frac{-4±\sqrt{16+468}}{2}$$

Tāpiri 16 ki te 468.

$$x=\frac{-4±\sqrt{484}}{2}$$

Tuhia te pūtakerua o te 484.

$$x=\frac{-4±22}{2}$$

Nā, me whakaoti te whārite x=\frac{-4±22}{2} ina he tāpiri te ±. Tāpiri -4 ki te 22.

$$x=\frac{18}{2}$$

Whakawehe 18 ki te 2.

$$x=9$$

Nā, me whakaoti te whārite x=\frac{-4±22}{2} ina he tango te ±. Tango 22 mai i -4.

$$x=-\frac{26}{2}$$

Whakawehe -26 ki te 2.

$$x=-13$$

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 9 mō te x_{1} me te -13 mō te x_{2}.

$$x^{2}+4x-117=\left(x-9\right)\left(x-\left(-13\right)\right)$$

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

$$x^{2}+4x-117=\left(x-9\right)\left(x+13\right)$$

Aromātai

$$\left(x-9\right)\left(x+13\right)$$