Whakaoti i te x^2+8x+16

$$x ^ { 2 } + 8 x + 16$$

Tauwehe

$$\left(x+4\right)^{2}$$

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.

$$a+b=8 ab=1\times 16=16$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga 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 8.

$$a=4 b=4$$

Tuhia anō te x^{2}+8x+16 hei \left(x^{2}+4x\right)+\left(4x+16\right).

$$\left(x^{2}+4x\right)+\left(4x+16\right)$$

Tauwehea te x i te tuatahi me te 4 i te rōpū tuarua.

$$x\left(x+4\right)+4\left(x+4\right)$$

Whakatauwehea atu te kīanga pātahi x+4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x+4\right)\left(x+4\right)$$

Tuhia anōtia hei pūrua huarua.

$$\left(x+4\right)^{2}$$

Ngā Upane Whakamahi Tauwehe o te Pūrua Huarua

Ko te tikanga tātai o tēnei huatoru he pūrua huatoru, ka whakareatia pea e tētahi tauwehe pātahi. Ka taea ngā pūrua huatoru te tauwehe mā te kimi i ngā pūtakerua o ngā kīanga tau ārahi, autō hoki.

$$factor(x^{2}+8x+16)$$

Kimihia te pūtakerua o te kīanga tau autō, 16.

$$\sqrt{16}=4$$

Ko te pūrua huatoru te pūrua o te huarua ko te tapeke tērā, te huatango rānei o ngā pūtakerua o ngā kīanga tau ārahi, autō hoki, e whakaritea ai te tohu e te tohu o te kīanga tau waenga o te pūrua huatoru.

$$\left(x+4\right)^{2}$$

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.

$$x^{2}+8x+16=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}-4\times 16}}{2}$$

Pūrua 8.

$$x=\frac{-8±\sqrt{64-4\times 16}}{2}$$

Whakareatia -4 ki te 16.

$$x=\frac{-8±\sqrt{64-64}}{2}$$

Tāpiri 64 ki te -64.

$$x=\frac{-8±\sqrt{0}}{2}$$

Tuhia te pūtakerua o te 0.

$$x=\frac{-8±0}{2}$$

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 -4 mō te x_{1} me te -4 mō te x_{2}.

$$x^{2}+8x+16=\left(x-\left(-4\right)\right)\left(x-\left(-4\right)\right)$$

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

$$x^{2}+8x+16=\left(x+4\right)\left(x+4\right)$$

Aromātai

$$\left(x+4\right)^{2}$$