Whakaoti i te x^2-2x-4

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

Tauwehe

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

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.

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

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.

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

Pūrua -2.

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

Whakareatia -4 ki te -4.

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

Tāpiri 4 ki te 16.

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

Tuhia te pūtakerua o te 20.

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

Ko te tauaro o -2 ko 2.

$$x=\frac{2±2\sqrt{5}}{2}$$

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

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

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

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

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

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

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

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

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

$$x^{2}-2x-4=\left(x-\left(\sqrt{5}+1\right)\right)\left(x-\left(1-\sqrt{5}\right)\right)$$

Aromātai

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