Whakaoti i te (x^2+3x+1)

$$( x ^ { 2 } + 3 x + 1 )$$

Tauwehe

$$\left(x-\frac{-\sqrt{5}-3}{2}\right)\left(x-\frac{\sqrt{5}-3}{2}\right)$$

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.

$$x=\frac{-3±\sqrt{3^{2}-4}}{2}$$

Pūrua 3.

$$x=\frac{-3±\sqrt{9-4}}{2}$$

Tāpiri 9 ki te -4.

$$x=\frac{-3±\sqrt{5}}{2}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{5}}{2} ina he tāpiri te ±. Tāpiri -3 ki te \sqrt{5}.

$$x=\frac{\sqrt{5}-3}{2}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{5}}{2} ina he tango te ±. Tango \sqrt{5} mai i -3.

$$x=\frac{-\sqrt{5}-3}{2}$$

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{5}}{2} mō te x_{1} me te \frac{-3-\sqrt{5}}{2} mō te x_{2}.

$$x^{2}+3x+1=\left(x-\frac{\sqrt{5}-3}{2}\right)\left(x-\frac{-\sqrt{5}-3}{2}\right)$$

Aromātai

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