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

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

Tauwehe

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

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 25b^{2}+pb+qb+4. Hei kimi p me q, whakaritea tētahi pūnaha kia whakaoti.

$$p+q=-20 pq=25\times 4=100$$

I te mea kua tōrunga te pq, he ōrite te tohu o p me q. I te mea kua tōraro te p+q, he tōraro hoki a p me q. 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.

$$p=-10 q=-10$$

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

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

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

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

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

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

Tuhia anōtia hei pūrua huarua.

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

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.

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

Kimihia te tauwehe pātahi nui rawa o ngā tau whakarea.

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

Kimihia te pūtakerua o te kīanga tau ārahi, 25b^{2}.

$$\sqrt{25b^{2}}=5b$$

Kimihia te pūtakerua o te kīanga tau autō, 4.

$$\sqrt{4}=2$$

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.

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

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.

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

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.

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

Pūrua -20.

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

Whakareatia -4 ki te 25.

$$b=\frac{-\left(-20\right)±\sqrt{400-100\times 4}}{2\times 25}$$

Whakareatia -100 ki te 4.

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

Tāpiri 400 ki te -400.

$$b=\frac{-\left(-20\right)±\sqrt{0}}{2\times 25}$$

Tuhia te pūtakerua o te 0.

$$b=\frac{-\left(-20\right)±0}{2\times 25}$$

Ko te tauaro o -20 ko 20.

$$b=\frac{20±0}{2\times 25}$$

Whakareatia 2 ki te 25.

$$b=\frac{20±0}{50}$$

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

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

Tango \frac{2}{5} mai i b 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.

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

Tango \frac{2}{5} mai i b 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.

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

Whakareatia \frac{5b-2}{5} ki te \frac{5b-2}{5} 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.

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

Whakareatia 5 ki te 5.

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

Whakakorea atu te tauwehe pūnoa nui rawa 25 i roto i te 25 me te 25.

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

Aromātai

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