Whakaoti i te =x^2-9x+1

$$= x ^ { 2 } - 9 x + 1$$

Tauwehe

$$\left(x-\frac{9-\sqrt{77}}{2}\right)\left(x-\frac{\sqrt{77}+9}{2}\right)$$

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.

$$x^{2}-9x+1=0$$

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.

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

Pūrua -9.

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

Tāpiri 81 ki te -4.

$$x=\frac{-\left(-9\right)±\sqrt{77}}{2}$$

Ko te tauaro o -9 ko 9.

$$x=\frac{9±\sqrt{77}}{2}$$

Nā, me whakaoti te whārite x=\frac{9±\sqrt{77}}{2} ina he tāpiri te ±. Tāpiri 9 ki te \sqrt{77}.

$$x=\frac{\sqrt{77}+9}{2}$$

Nā, me whakaoti te whārite x=\frac{9±\sqrt{77}}{2} ina he tango te ±. Tango \sqrt{77} mai i 9.

$$x=\frac{9-\sqrt{77}}{2}$$

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{77}}{2} mō te x_{1} me te \frac{9-\sqrt{77}}{2} mō te x_{2}.

$$x^{2}-9x+1=\left(x-\frac{\sqrt{77}+9}{2}\right)\left(x-\frac{9-\sqrt{77}}{2}\right)$$

Aromātai

$$x^{2}-9x+1$$