Whakaoti i te 2x^2-2x-2

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

Tauwehe

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

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.

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

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.

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

Pūrua -2.

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

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-2\right)±\sqrt{4-8\left(-2\right)}}{2\times 2}$$

Whakareatia -8 ki te -2.

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

Tāpiri 4 ki te 16.

$$x=\frac{-\left(-2\right)±\sqrt{20}}{2\times 2}$$

Tuhia te pūtakerua o te 20.

$$x=\frac{-\left(-2\right)±2\sqrt{5}}{2\times 2}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±2\sqrt{5}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{2±2\sqrt{5}}{4}$$

Nā, me whakaoti te whārite x=\frac{2±2\sqrt{5}}{4} ina he tāpiri te ±. Tāpiri 2 ki te 2\sqrt{5}.

$$x=\frac{2\sqrt{5}+2}{4}$$

Whakawehe 2+2\sqrt{5} ki te 4.

$$x=\frac{\sqrt{5}+1}{2}$$

Nā, me whakaoti te whārite x=\frac{2±2\sqrt{5}}{4} ina he tango te ±. Tango 2\sqrt{5} mai i 2.

$$x=\frac{2-2\sqrt{5}}{4}$$

Whakawehe 2-2\sqrt{5} ki te 4.

$$x=\frac{1-\sqrt{5}}{2}$$

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

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

Aromātai

$$2\left(x^{2}-x-1\right)$$