Whakaoti i te x^2-9x-10

$${ x }^{ 2 } -9x-10$$

Tauwehe

$$\left(x-10\right)\left(x+1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx-10. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-9 ab=1\left(-10\right)=-10$$

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 -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.

$$a=-10 b=1$$

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

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

Whakatauwehea atu x i te x^{2}-10x.

$$x\left(x-10\right)+x-10$$

Whakatauwehea atu te kīanga pātahi x-10 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-10\right)\left(x+1\right)$$

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.

$$x^{2}-9x-10=0$$

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.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-10\right)}}{2}$$

Pūrua -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\left(-10\right)}}{2}$$

Whakareatia -4 ki te -10.

$$x=\frac{-\left(-9\right)±\sqrt{81+40}}{2}$$

Tāpiri 81 ki te 40.

$$x=\frac{-\left(-9\right)±\sqrt{121}}{2}$$

Tuhia te pūtakerua o te 121.

$$x=\frac{-\left(-9\right)±11}{2}$$

Ko te tauaro o -9 ko 9.

$$x=\frac{9±11}{2}$$

Nā, me whakaoti te whārite x=\frac{9±11}{2} ina he tāpiri te ±. Tāpiri 9 ki te 11.

$$x=\frac{20}{2}$$

Whakawehe 20 ki te 2.

$$x=10$$

Nā, me whakaoti te whārite x=\frac{9±11}{2} ina he tango te ±. Tango 11 mai i 9.

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

Whakawehe -2 ki te 2.

$$x=-1$$

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 10 mō te x_{1} me te -1 mō te x_{2}.

$$x^{2}-9x-10=\left(x-10\right)\left(x-\left(-1\right)\right)$$

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

$$x^{2}-9x-10=\left(x-10\right)\left(x+1\right)$$

Aromātai

$$\left(x-10\right)\left(x+1\right)$$