Whakaoti i te 3x^2+5x+1?

$$3 x ^ { 2 } + 5 x + 1 ?$$

Tauwehe

$$3\left(x-\frac{-\sqrt{13}-5}{6}\right)\left(x-\frac{\sqrt{13}-5}{6}\right)$$

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.

$$3x^{2}+5x+1=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\times 3}}{2\times 3}$$

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4\times 3}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-5±\sqrt{25-12}}{2\times 3}$$

Tāpiri 25 ki te -12.

$$x=\frac{-5±\sqrt{13}}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-5±\sqrt{13}}{6}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{13}}{6} ina he tāpiri te ±. Tāpiri -5 ki te \sqrt{13}.

$$x=\frac{\sqrt{13}-5}{6}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{13}}{6} ina he tango te ±. Tango \sqrt{13} mai i -5.

$$x=\frac{-\sqrt{13}-5}{6}$$

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{-5+\sqrt{13}}{6} mō te x_{1} me te \frac{-5-\sqrt{13}}{6} mō te x_{2}.

$$3x^{2}+5x+1=3\left(x-\frac{\sqrt{13}-5}{6}\right)\left(x-\frac{-\sqrt{13}-5}{6}\right)$$

Aromātai

$$3x^{2}+5x+1$$