Whakaoti i te (x^2-8x-4)

$$( x ^ { 2 } - 8 x - 4 )$$

Tauwehe

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

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.

$$x^{2}-8x-4=0$$

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.

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

Pūrua -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-4\right)}}{2}$$

Whakareatia -4 ki te -4.

$$x=\frac{-\left(-8\right)±\sqrt{64+16}}{2}$$

Tāpiri 64 ki te 16.

$$x=\frac{-\left(-8\right)±\sqrt{80}}{2}$$

Tuhia te pūtakerua o te 80.

$$x=\frac{-\left(-8\right)±4\sqrt{5}}{2}$$

Ko te tauaro o -8 ko 8.

$$x=\frac{8±4\sqrt{5}}{2}$$

Nā, me whakaoti te whārite x=\frac{8±4\sqrt{5}}{2} ina he tāpiri te ±. Tāpiri 8 ki te 4\sqrt{5}.

$$x=\frac{4\sqrt{5}+8}{2}$$

Whakawehe 8+4\sqrt{5} ki te 2.

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

Nā, me whakaoti te whārite x=\frac{8±4\sqrt{5}}{2} ina he tango te ±. Tango 4\sqrt{5} mai i 8.

$$x=\frac{8-4\sqrt{5}}{2}$$

Whakawehe 8-4\sqrt{5} ki te 2.

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

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 4+2\sqrt{5} mō te x_{1} me te 4-2\sqrt{5} mō te x_{2}.

$$x^{2}-8x-4=\left(x-\left(2\sqrt{5}+4\right)\right)\left(x-\left(4-2\sqrt{5}\right)\right)$$

Aromātai

$$x^{2}-8x-4$$