Whakaoti i te x^2-x-3

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

Tauwehe

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

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.

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

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.

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

Whakareatia -4 ki te -3.

$$x=\frac{-\left(-1\right)±\sqrt{1+12}}{2}$$

Tāpiri 1 ki te 12.

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

Ko te tauaro o -1 ko 1.

$$x=\frac{1±\sqrt{13}}{2}$$

Nā, me whakaoti te whārite x=\frac{1±\sqrt{13}}{2} ina he tāpiri te ±. Tāpiri 1 ki te \sqrt{13}.

$$x=\frac{\sqrt{13}+1}{2}$$

Nā, me whakaoti te whārite x=\frac{1±\sqrt{13}}{2} ina he tango te ±. Tango \sqrt{13} mai i 1.

$$x=\frac{1-\sqrt{13}}{2}$$

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

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

Aromātai

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