Whakaoti i te x^2-10x+16

$$x ^ { 2 } - 10 x + 16$$

Tauwehe

$$\left(x-8\right)\left(x-2\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx+16. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-10 ab=1\times 16=16$$

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 16.

$$-1,-16 -2,-8 -4,-4$$

Tātaihia te tapeke mō ia takirua.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

Ko te otinga te takirua ka hoatu i te tapeke -10.

$$a=-8 b=-2$$

Tuhia anō te x^{2}-10x+16 hei \left(x^{2}-8x\right)+\left(-2x+16\right).

$$\left(x^{2}-8x\right)+\left(-2x+16\right)$$

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

$$x\left(x-8\right)-2\left(x-8\right)$$

Whakatauwehea atu te kīanga pātahi x-8 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-8\right)\left(x-2\right)$$

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.

$$x^{2}-10x+16=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 16}}{2}$$

Pūrua -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 16}}{2}$$

Whakareatia -4 ki te 16.

$$x=\frac{-\left(-10\right)±\sqrt{100-64}}{2}$$

Tāpiri 100 ki te -64.

$$x=\frac{-\left(-10\right)±\sqrt{36}}{2}$$

Tuhia te pūtakerua o te 36.

$$x=\frac{-\left(-10\right)±6}{2}$$

Ko te tauaro o -10 ko 10.

$$x=\frac{10±6}{2}$$

Nā, me whakaoti te whārite x=\frac{10±6}{2} ina he tāpiri te ±. Tāpiri 10 ki te 6.

$$x=\frac{16}{2}$$

Whakawehe 16 ki te 2.

$$x=8$$

Nā, me whakaoti te whārite x=\frac{10±6}{2} ina he tango te ±. Tango 6 mai i 10.

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

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 8 mō te x_{1} me te 2 mō te x_{2}.

$$x^{2}-10x+16=\left(x-8\right)\left(x-2\right)$$

Aromātai

$$\left(x-8\right)\left(x-2\right)$$