Whakaoti i te 2x^2-9x+5

$$2 x ^ { 2 } - 9 x + 5$$

Tauwehe

$$2\left(x-\frac{9-\sqrt{41}}{4}\right)\left(x-\frac{\sqrt{41}+9}{4}\right)$$

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.

$$2x^{2}-9x+5=0$$

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.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 2\times 5}}{2\times 2}$$

Pūrua -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 2\times 5}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-9\right)±\sqrt{81-8\times 5}}{2\times 2}$$

Whakareatia -8 ki te 5.

$$x=\frac{-\left(-9\right)±\sqrt{81-40}}{2\times 2}$$

Tāpiri 81 ki te -40.

$$x=\frac{-\left(-9\right)±\sqrt{41}}{2\times 2}$$

Ko te tauaro o -9 ko 9.

$$x=\frac{9±\sqrt{41}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{9±\sqrt{41}}{4}$$

Nā, me whakaoti te whārite x=\frac{9±\sqrt{41}}{4} ina he tāpiri te ±. Tāpiri 9 ki te \sqrt{41}.

$$x=\frac{\sqrt{41}+9}{4}$$

Nā, me whakaoti te whārite x=\frac{9±\sqrt{41}}{4} ina he tango te ±. Tango \sqrt{41} mai i 9.

$$x=\frac{9-\sqrt{41}}{4}$$

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{9+\sqrt{41}}{4} mō te x_{1} me te \frac{9-\sqrt{41}}{4} mō te x_{2}.

$$2x^{2}-9x+5=2\left(x-\frac{\sqrt{41}+9}{4}\right)\left(x-\frac{9-\sqrt{41}}{4}\right)$$

Aromātai

$$2x^{2}-9x+5$$