Whakaoti i te x^2-14x+45

$$x ^ { 2 } - 14 x + 45$$

Tauwehe

$$\left(x-9\right)\left(x-5\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx+45. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-14 ab=1\times 45=45$$

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 45.

$$-1,-45 -3,-15 -5,-9$$

Tātaihia te tapeke mō ia takirua.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

Ko te otinga te takirua ka hoatu i te tapeke -14.

$$a=-9 b=-5$$

Tuhia anō te x^{2}-14x+45 hei \left(x^{2}-9x\right)+\left(-5x+45\right).

$$\left(x^{2}-9x\right)+\left(-5x+45\right)$$

Tauwehea te x i te tuatahi me te -5 i te rōpū tuarua.

$$x\left(x-9\right)-5\left(x-9\right)$$

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

$$\left(x-9\right)\left(x-5\right)$$

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.

$$x^{2}-14x+45=0$$

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.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 45}}{2}$$

Pūrua -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 45}}{2}$$

Whakareatia -4 ki te 45.

$$x=\frac{-\left(-14\right)±\sqrt{196-180}}{2}$$

Tāpiri 196 ki te -180.

$$x=\frac{-\left(-14\right)±\sqrt{16}}{2}$$

Tuhia te pūtakerua o te 16.

$$x=\frac{-\left(-14\right)±4}{2}$$

Ko te tauaro o -14 ko 14.

$$x=\frac{14±4}{2}$$

Nā, me whakaoti te whārite x=\frac{14±4}{2} ina he tāpiri te ±. Tāpiri 14 ki te 4.

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

Whakawehe 18 ki te 2.

$$x=9$$

Nā, me whakaoti te whārite x=\frac{14±4}{2} ina he tango te ±. Tango 4 mai i 14.

$$x=\frac{10}{2}$$

Whakawehe 10 ki te 2.

$$x=5$$

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 5 mō te x_{2}.

$$x^{2}-14x+45=\left(x-9\right)\left(x-5\right)$$

Aromātai

$$\left(x-9\right)\left(x-5\right)$$