Whakaoti i te f(x)=3x^2+4x-2

$$f ( x ) = 3 x ^ { 2 } + 4 x - 2$$

Tauwehe

$$3\left(x-\frac{-\sqrt{10}-2}{3}\right)\left(x-\frac{\sqrt{10}-2}{3}\right)$$

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.

$$3x^{2}+4x-2=0$$

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.

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

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\times 3\left(-2\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-4±\sqrt{16-12\left(-2\right)}}{2\times 3}$$

Whakareatia -12 ki te -2.

$$x=\frac{-4±\sqrt{16+24}}{2\times 3}$$

Tāpiri 16 ki te 24.

$$x=\frac{-4±\sqrt{40}}{2\times 3}$$

Tuhia te pūtakerua o te 40.

$$x=\frac{-4±2\sqrt{10}}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-4±2\sqrt{10}}{6}$$

Nā, me whakaoti te whārite x=\frac{-4±2\sqrt{10}}{6} ina he tāpiri te ±. Tāpiri -4 ki te 2\sqrt{10}.

$$x=\frac{2\sqrt{10}-4}{6}$$

Whakawehe -4+2\sqrt{10} ki te 6.

$$x=\frac{\sqrt{10}-2}{3}$$

Nā, me whakaoti te whārite x=\frac{-4±2\sqrt{10}}{6} ina he tango te ±. Tango 2\sqrt{10} mai i -4.

$$x=\frac{-2\sqrt{10}-4}{6}$$

Whakawehe -4-2\sqrt{10} ki te 6.

$$x=\frac{-\sqrt{10}-2}{3}$$

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{-2+\sqrt{10}}{3} mō te x_{1} me te \frac{-2-\sqrt{10}}{3} mō te x_{2}.

$$3x^{2}+4x-2=3\left(x-\frac{\sqrt{10}-2}{3}\right)\left(x-\frac{-\sqrt{10}-2}{3}\right)$$

Aromātai

$$3x^{2}+4x-2$$