Whakaoti i te 25b^2-10b

$$25 b ^ { 2 } - 10 b$$

Tauwehe

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

Ngā Upane Otinga

Tauwehea te 5.

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

Whakaarohia te 5b^{2}-2b. Tauwehea te b.

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

Me tuhi anō te kīanga whakatauwehe katoa.

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

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.

$$25b^{2}-10b=0$$

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.

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

Tuhia te pūtakerua o te \left(-10\right)^{2}.

$$b=\frac{-\left(-10\right)±10}{2\times 25}$$

Ko te tauaro o -10 ko 10.

$$b=\frac{10±10}{2\times 25}$$

Whakareatia 2 ki te 25.

$$b=\frac{10±10}{50}$$

Nā, me whakaoti te whārite b=\frac{10±10}{50} ina he tāpiri te ±. Tāpiri 10 ki te 10.

$$b=\frac{20}{50}$$

Whakahekea te hautanga \frac{20}{50} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

$$b=\frac{2}{5}$$

Nā, me whakaoti te whārite b=\frac{10±10}{50} ina he tango te ±. Tango 10 mai i 10.

$$b=\frac{0}{50}$$

Whakawehe 0 ki te 50.

$$b=0$$

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 0 mō te x_{2}.

$$25b^{2}-10b=25\left(b-\frac{2}{5}\right)b$$

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.

$$25b^{2}-10b=25\times \frac{5b-2}{5}b$$

Whakakorea atu te tauwehe pūnoa nui rawa 5 i roto i te 25 me te 5.

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

Aromātai

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