Whakaoti i te 25-10x+x^2

$$25 - 10 x + x ^ { 2 }$$

Tauwehe

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

Ngā Upane Otinga

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$x^{2}-10x+25$$

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx+25. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-10 ab=1\times 25=25$$

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 25.

$$-1,-25 -5,-5$$

Tātaihia te tapeke mō ia takirua.

$$-1-25=-26 -5-5=-10$$

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

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

Tuhia anō te x^{2}-10x+25 hei \left(x^{2}-5x\right)+\left(-5x+25\right).

$$\left(x^{2}-5x\right)+\left(-5x+25\right)$$

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

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

Whakatauwehea atu te kīanga pātahi x-5 mā te whakamahi i te āhuatanga tātai tohatoha.

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

Tuhia anōtia hei pūrua huarua.

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

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.

$$factor(x^{2}-10x+25)$$

Kimihia te pūtakerua o te kīanga tau autō, 25.

$$\sqrt{25}=5$$

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.

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

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.

$$x^{2}-10x+25=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 25}}{2}$$

Pūrua -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 25}}{2}$$

Whakareatia -4 ki te 25.

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

Tāpiri 100 ki te -100.

$$x=\frac{-\left(-10\right)±\sqrt{0}}{2}$$

Tuhia te pūtakerua o te 0.

$$x=\frac{-\left(-10\right)±0}{2}$$

Ko te tauaro o -10 ko 10.

$$x=\frac{10±0}{2}$$

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 5 mō te x_{1} me te 5 mō te x_{2}.

$$x^{2}-10x+25=\left(x-5\right)\left(x-5\right)$$

Aromātai

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