Whakaoti i te x^2-8x+5

$$x ^ { 2 } - 8 x + 5$$

Tauwehe

$$\left(x-\left(4-\sqrt{11}\right)\right)\left(x-\left(\sqrt{11}+4\right)\right)$$

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.

$$x^{2}-8x+5=0$$

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.

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

Pūrua -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 5}}{2}$$

Whakareatia -4 ki te 5.

$$x=\frac{-\left(-8\right)±\sqrt{64-20}}{2}$$

Tāpiri 64 ki te -20.

$$x=\frac{-\left(-8\right)±\sqrt{44}}{2}$$

Tuhia te pūtakerua o te 44.

$$x=\frac{-\left(-8\right)±2\sqrt{11}}{2}$$

Ko te tauaro o -8 ko 8.

$$x=\frac{8±2\sqrt{11}}{2}$$

Nā, me whakaoti te whārite x=\frac{8±2\sqrt{11}}{2} ina he tāpiri te ±. Tāpiri 8 ki te 2\sqrt{11}.

$$x=\frac{2\sqrt{11}+8}{2}$$

Whakawehe 8+2\sqrt{11} ki te 2.

$$x=\sqrt{11}+4$$

Nā, me whakaoti te whārite x=\frac{8±2\sqrt{11}}{2} ina he tango te ±. Tango 2\sqrt{11} mai i 8.

$$x=\frac{8-2\sqrt{11}}{2}$$

Whakawehe 8-2\sqrt{11} ki te 2.

$$x=4-\sqrt{11}$$

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+\sqrt{11} mō te x_{1} me te 4-\sqrt{11} mō te x_{2}.

$$x^{2}-8x+5=\left(x-\left(\sqrt{11}+4\right)\right)\left(x-\left(4-\sqrt{11}\right)\right)$$

Aromātai

$$x^{2}-8x+5$$