Whakaoti i te b^2-5*b=

$$b ^ { 2 } - 5 \cdot b =$$

Tauwehe

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

Ngā Upane Otinga

Tauwehea te b.

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

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.

$$b^{2}-5b=0$$

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.

$$b=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}}}{2}$$

Tuhia te pūtakerua o te \left(-5\right)^{2}.

$$b=\frac{-\left(-5\right)±5}{2}$$

Ko te tauaro o -5 ko 5.

$$b=\frac{5±5}{2}$$

Nā, me whakaoti te whārite b=\frac{5±5}{2} ina he tāpiri te ±. Tāpiri 5 ki te 5.

$$b=\frac{10}{2}$$

Whakawehe 10 ki te 2.

$$b=5$$

Nā, me whakaoti te whārite b=\frac{5±5}{2} ina he tango te ±. Tango 5 mai i 5.

$$b=\frac{0}{2}$$

Whakawehe 0 ki te 2.

$$b=0$$

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 5 mō te x_{1} me te 0 mō te x_{2}.

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

Aromātai

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