Whakaoti i te -x^2-2x+4

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

Tauwehe

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

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.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Pūrua -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$x=\frac{-\left(-2\right)±\sqrt{4+4\times 4}}{2\left(-1\right)}$$

Whakareatia 4 ki te 4.

$$x=\frac{-\left(-2\right)±\sqrt{4+16}}{2\left(-1\right)}$$

Tāpiri 4 ki te 16.

$$x=\frac{-\left(-2\right)±\sqrt{20}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te 20.

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

Ko te tauaro o -2 ko 2.

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

Whakareatia 2 ki te -1.

$$x=\frac{2±2\sqrt{5}}{-2}$$

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.

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

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.

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

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 -\left(1+\sqrt{5}\right) mō te x_{1} me te -1+\sqrt{5} mō te x_{2}.

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

Aromātai

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