Whakaoti i te x^2-11x-26

$${ x }^{ 2 } -11x-26$$

Tauwehe

$$\left(x-13\right)\left(x+2\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx-26. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-11 ab=1\left(-26\right)=-26$$

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 -26.

$$1,-26 2,-13$$

Tātaihia te tapeke mō ia takirua.

$$1-26=-25 2-13=-11$$

Ko te otinga te takirua ka hoatu i te tapeke -11.

$$a=-13 b=2$$

Tuhia anō te x^{2}-11x-26 hei \left(x^{2}-13x\right)+\left(2x-26\right).

$$\left(x^{2}-13x\right)+\left(2x-26\right)$$

Tauwehea te x i te tuatahi me te 2 i te rōpū tuarua.

$$x\left(x-13\right)+2\left(x-13\right)$$

Whakatauwehea atu te kīanga pātahi x-13 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-13\right)\left(x+2\right)$$

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.

$$x^{2}-11x-26=0$$

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.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-26\right)}}{2}$$

Pūrua -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\left(-26\right)}}{2}$$

Whakareatia -4 ki te -26.

$$x=\frac{-\left(-11\right)±\sqrt{121+104}}{2}$$

Tāpiri 121 ki te 104.

$$x=\frac{-\left(-11\right)±\sqrt{225}}{2}$$

Tuhia te pūtakerua o te 225.

$$x=\frac{-\left(-11\right)±15}{2}$$

Ko te tauaro o -11 ko 11.

$$x=\frac{11±15}{2}$$

Nā, me whakaoti te whārite x=\frac{11±15}{2} ina he tāpiri te ±. Tāpiri 11 ki te 15.

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

Whakawehe 26 ki te 2.

$$x=13$$

Nā, me whakaoti te whārite x=\frac{11±15}{2} ina he tango te ±. Tango 15 mai i 11.

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

Whakawehe -4 ki te 2.

$$x=-2$$

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 13 mō te x_{1} me te -2 mō te x_{2}.

$$x^{2}-11x-26=\left(x-13\right)\left(x-\left(-2\right)\right)$$

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

$$x^{2}-11x-26=\left(x-13\right)\left(x+2\right)$$

Aromātai

$$\left(x-13\right)\left(x+2\right)$$