Whakaoti i te f(x)=1-3x-x^2

$$f ( x ) = 1 - 3 x - x ^ { 2 }$$

Tauwehe

$$-\left(x-\frac{-\sqrt{13}-3}{2}\right)\left(x-\frac{\sqrt{13}-3}{2}\right)$$

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.

$$-x^{2}-3x+1=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)}}{2\left(-1\right)}$$

Pūrua -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+4}}{2\left(-1\right)}$$

Tāpiri 9 ki te 4.

$$x=\frac{-\left(-3\right)±\sqrt{13}}{2\left(-1\right)}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±\sqrt{13}}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{3±\sqrt{13}}{-2}$$

Nā, me whakaoti te whārite x=\frac{3±\sqrt{13}}{-2} ina he tāpiri te ±. Tāpiri 3 ki te \sqrt{13}.

$$x=\frac{\sqrt{13}+3}{-2}$$

Whakawehe 3+\sqrt{13} ki te -2.

$$x=\frac{-\sqrt{13}-3}{2}$$

Nā, me whakaoti te whārite x=\frac{3±\sqrt{13}}{-2} ina he tango te ±. Tango \sqrt{13} mai i 3.

$$x=\frac{3-\sqrt{13}}{-2}$$

Whakawehe 3-\sqrt{13} ki te -2.

$$x=\frac{\sqrt{13}-3}{2}$$

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{-3-\sqrt{13}}{2} mō te x_{1} me te \frac{-3+\sqrt{13}}{2} mō te x_{2}.

$$-x^{2}-3x+1=-\left(x-\frac{-\sqrt{13}-3}{2}\right)\left(x-\frac{\sqrt{13}-3}{2}\right)$$

Aromātai

$$1-3x-x^{2}$$