Whakaoti i te x^2-19x+90

$$x ^ { 2 } - 19 x + 90$$

Tauwehe

$$\left(x-10\right)\left(x-9\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx+90. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-19 ab=1\times 90=90$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 90.

$$-1,-90 -2,-45 -3,-30 -5,-18 -6,-15 -9,-10$$

Tātaihia te tapeke mō ia takirua.

$$-1-90=-91 -2-45=-47 -3-30=-33 -5-18=-23 -6-15=-21 -9-10=-19$$

Ko te otinga te takirua ka hoatu i te tapeke -19.

$$a=-10 b=-9$$

Tuhia anō te x^{2}-19x+90 hei \left(x^{2}-10x\right)+\left(-9x+90\right).

$$\left(x^{2}-10x\right)+\left(-9x+90\right)$$

Tauwehea te x i te tuatahi me te -9 i te rōpū tuarua.

$$x\left(x-10\right)-9\left(x-10\right)$$

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

$$\left(x-10\right)\left(x-9\right)$$

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.

$$x^{2}-19x+90=0$$

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.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 90}}{2}$$

Pūrua -19.

$$x=\frac{-\left(-19\right)±\sqrt{361-4\times 90}}{2}$$

Whakareatia -4 ki te 90.

$$x=\frac{-\left(-19\right)±\sqrt{361-360}}{2}$$

Tāpiri 361 ki te -360.

$$x=\frac{-\left(-19\right)±\sqrt{1}}{2}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-19\right)±1}{2}$$

Ko te tauaro o -19 ko 19.

$$x=\frac{19±1}{2}$$

Nā, me whakaoti te whārite x=\frac{19±1}{2} ina he tāpiri te ±. Tāpiri 19 ki te 1.

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

Whakawehe 20 ki te 2.

$$x=10$$

Nā, me whakaoti te whārite x=\frac{19±1}{2} ina he tango te ±. Tango 1 mai i 19.

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

Whakawehe 18 ki te 2.

$$x=9$$

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 9 mō te x_{2}.

$$x^{2}-19x+90=\left(x-10\right)\left(x-9\right)$$

Aromātai

$$\left(x-10\right)\left(x-9\right)$$