Whakaoti i te x^2-12x+35

$$x ^ { 2 } - 12 x + 35$$

Tauwehe

$$\left(x-7\right)\left(x-5\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx+35. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-12 ab=1\times 35=35$$

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 35.

$$-1,-35 -5,-7$$

Tātaihia te tapeke mō ia takirua.

$$-1-35=-36 -5-7=-12$$

Ko te otinga te takirua ka hoatu i te tapeke -12.

$$a=-7 b=-5$$

Tuhia anō te x^{2}-12x+35 hei \left(x^{2}-7x\right)+\left(-5x+35\right).

$$\left(x^{2}-7x\right)+\left(-5x+35\right)$$

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

$$x\left(x-7\right)-5\left(x-7\right)$$

Whakatauwehea atu te kīanga pātahi x-7 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-7\right)\left(x-5\right)$$

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.

$$x^{2}-12x+35=0$$

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.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 35}}{2}$$

Pūrua -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 35}}{2}$$

Whakareatia -4 ki te 35.

$$x=\frac{-\left(-12\right)±\sqrt{144-140}}{2}$$

Tāpiri 144 ki te -140.

$$x=\frac{-\left(-12\right)±\sqrt{4}}{2}$$

Tuhia te pūtakerua o te 4.

$$x=\frac{-\left(-12\right)±2}{2}$$

Ko te tauaro o -12 ko 12.

$$x=\frac{12±2}{2}$$

Nā, me whakaoti te whārite x=\frac{12±2}{2} ina he tāpiri te ±. Tāpiri 12 ki te 2.

$$x=\frac{14}{2}$$

Whakawehe 14 ki te 2.

$$x=7$$

Nā, me whakaoti te whārite x=\frac{12±2}{2} ina he tango te ±. Tango 2 mai i 12.

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

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 7 mō te x_{1} me te 5 mō te x_{2}.

$$x^{2}-12x+35=\left(x-7\right)\left(x-5\right)$$

Aromātai

$$\left(x-7\right)\left(x-5\right)$$