Whakaoti i te 25-20r+4r^2

$$25 - 20 r + 4 r ^ { 2 }$$

Tauwehe

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

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.

$$4r^{2}-20r+25$$

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 4r^{2}+ar+br+25. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-20 ab=4\times 25=100$$

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 100.

$$-1,-100 -2,-50 -4,-25 -5,-20 -10,-10$$

Tātaihia te tapeke mō ia takirua.

$$-1-100=-101 -2-50=-52 -4-25=-29 -5-20=-25 -10-10=-20$$

Ko te otinga te takirua ka hoatu i te tapeke -20.

$$a=-10 b=-10$$

Tuhia anō te 4r^{2}-20r+25 hei \left(4r^{2}-10r\right)+\left(-10r+25\right).

$$\left(4r^{2}-10r\right)+\left(-10r+25\right)$$

Tauwehea te 2r i te tuatahi me te -5 i te rōpū tuarua.

$$2r\left(2r-5\right)-5\left(2r-5\right)$$

Whakatauwehea atu te kīanga pātahi 2r-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2r-5\right)\left(2r-5\right)$$

Tuhia anōtia hei pūrua huarua.

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

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.

$$factor(4r^{2}-20r+25)$$

Kimihia te tauwehe pātahi nui rawa o ngā tau whakarea.

$$gcf(4,-20,25)=1$$

Kimihia te pūtakerua o te kīanga tau ārahi, 4r^{2}.

$$\sqrt{4r^{2}}=2r$$

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.

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

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.

$$4r^{2}-20r+25=0$$

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.

$$r=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 4\times 25}}{2\times 4}$$

Pūrua -20.

$$r=\frac{-\left(-20\right)±\sqrt{400-4\times 4\times 25}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$r=\frac{-\left(-20\right)±\sqrt{400-16\times 25}}{2\times 4}$$

Whakareatia -16 ki te 25.

$$r=\frac{-\left(-20\right)±\sqrt{400-400}}{2\times 4}$$

Tāpiri 400 ki te -400.

$$r=\frac{-\left(-20\right)±\sqrt{0}}{2\times 4}$$

Tuhia te pūtakerua o te 0.

$$r=\frac{-\left(-20\right)±0}{2\times 4}$$

Ko te tauaro o -20 ko 20.

$$r=\frac{20±0}{2\times 4}$$

Whakareatia 2 ki te 4.

$$r=\frac{20±0}{8}$$

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 \frac{5}{2} mō te x_{1} me te \frac{5}{2} mō te x_{2}.

$$4r^{2}-20r+25=4\left(r-\frac{5}{2}\right)\left(r-\frac{5}{2}\right)$$

Tango \frac{5}{2} mai i r mā te kimi i te tauraro pātahi me te tango i ngā taurunga, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$4r^{2}-20r+25=4\times \frac{2r-5}{2}\left(r-\frac{5}{2}\right)$$

Tango \frac{5}{2} mai i r mā te kimi i te tauraro pātahi me te tango i ngā taurunga, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$4r^{2}-20r+25=4\times \frac{2r-5}{2}\times \frac{2r-5}{2}$$

Whakareatia \frac{2r-5}{2} ki te \frac{2r-5}{2} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$4r^{2}-20r+25=4\times \frac{\left(2r-5\right)\left(2r-5\right)}{2\times 2}$$

Whakareatia 2 ki te 2.

$$4r^{2}-20r+25=4\times \frac{\left(2r-5\right)\left(2r-5\right)}{4}$$

Whakakorea atu te tauwehe pūnoa nui rawa 4 i roto i te 4 me te 4.

$$4r^{2}-20r+25=\left(2r-5\right)\left(2r-5\right)$$

Aromātai

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