Whakaoti i te +x^2-5x

$$+ x ^ { 2 } - 5 x$$

Tauwehe

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

Ngā Upane Otinga

Tauwehea te x.

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

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.

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

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.

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

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

$$x=\frac{-\left(-5\right)±5}{2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±5}{2}$$

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

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

Whakawehe 10 ki te 2.

$$x=5$$

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

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

Whakawehe 0 ki te 2.

$$x=0$$

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}.

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

Aromātai

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