Whakaoti i te 3x^2-9x+1

$$3 { x }^{ 2 } -9x+1$$

Tauwehe

$$3\left(x-\left(-\frac{\sqrt{69}}{6}+\frac{3}{2}\right)\right)\left(x-\left(\frac{\sqrt{69}}{6}+\frac{3}{2}\right)\right)$$

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.

$$3x^{2}-9x+1=0$$

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.

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

Pūrua -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 3}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-9\right)±\sqrt{81-12}}{2\times 3}$$

Tāpiri 81 ki te -12.

$$x=\frac{-\left(-9\right)±\sqrt{69}}{2\times 3}$$

Ko te tauaro o -9 ko 9.

$$x=\frac{9±\sqrt{69}}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{9±\sqrt{69}}{6}$$

Nā, me whakaoti te whārite x=\frac{9±\sqrt{69}}{6} ina he tāpiri te ±. Tāpiri 9 ki te \sqrt{69}.

$$x=\frac{\sqrt{69}+9}{6}$$

Whakawehe 9+\sqrt{69} ki te 6.

$$x=\frac{\sqrt{69}}{6}+\frac{3}{2}$$

Nā, me whakaoti te whārite x=\frac{9±\sqrt{69}}{6} ina he tango te ±. Tango \sqrt{69} mai i 9.

$$x=\frac{9-\sqrt{69}}{6}$$

Whakawehe 9-\sqrt{69} ki te 6.

$$x=-\frac{\sqrt{69}}{6}+\frac{3}{2}$$

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}{2}+\frac{\sqrt{69}}{6} mō te x_{1} me te \frac{3}{2}-\frac{\sqrt{69}}{6} mō te x_{2}.

$$3x^{2}-9x+1=3\left(x-\left(\frac{\sqrt{69}}{6}+\frac{3}{2}\right)\right)\left(x-\left(-\frac{\sqrt{69}}{6}+\frac{3}{2}\right)\right)$$

Aromātai

$$3x^{2}-9x+1$$