Whakaoti i te f(x)=x^2-4x+1

$$f ( x ) = x ^ { 2 } - 4 x + 1$$

Tauwehe

$$\left(x-\left(2-\sqrt{3}\right)\right)\left(x-\left(\sqrt{3}+2\right)\right)$$

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.

$$x^{2}-4x+1=0$$

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.

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

Pūrua -4.

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

Tāpiri 16 ki te -4.

$$x=\frac{-\left(-4\right)±\sqrt{12}}{2}$$

Tuhia te pūtakerua o te 12.

$$x=\frac{-\left(-4\right)±2\sqrt{3}}{2}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±2\sqrt{3}}{2}$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{3}}{2} ina he tāpiri te ±. Tāpiri 4 ki te 2\sqrt{3}.

$$x=\frac{2\sqrt{3}+4}{2}$$

Whakawehe 4+2\sqrt{3} ki te 2.

$$x=\sqrt{3}+2$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{3}}{2} ina he tango te ±. Tango 2\sqrt{3} mai i 4.

$$x=\frac{4-2\sqrt{3}}{2}$$

Whakawehe 4-2\sqrt{3} ki te 2.

$$x=2-\sqrt{3}$$

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 2+\sqrt{3} mō te x_{1} me te 2-\sqrt{3} mō te x_{2}.

$$x^{2}-4x+1=\left(x-\left(\sqrt{3}+2\right)\right)\left(x-\left(2-\sqrt{3}\right)\right)$$

Aromātai

$$x^{2}-4x+1$$