Whakaoti i te g(x)=x^2-5x+2

$$g ( x ) = x ^ { 2 } - 5 x + 2$$

Tauwehe

$$\left(x-\frac{5-\sqrt{17}}{2}\right)\left(x-\frac{\sqrt{17}+5}{2}\right)$$

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.

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

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.

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

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2}}{2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8}}{2}$$

Tāpiri 25 ki te -8.

$$x=\frac{-\left(-5\right)±\sqrt{17}}{2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±\sqrt{17}}{2}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{17}}{2} ina he tāpiri te ±. Tāpiri 5 ki te \sqrt{17}.

$$x=\frac{\sqrt{17}+5}{2}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{17}}{2} ina he tango te ±. Tango \sqrt{17} mai i 5.

$$x=\frac{5-\sqrt{17}}{2}$$

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 \frac{5+\sqrt{17}}{2} mō te x_{1} me te \frac{5-\sqrt{17}}{2} mō te x_{2}.

$$x^{2}-5x+2=\left(x-\frac{\sqrt{17}+5}{2}\right)\left(x-\frac{5-\sqrt{17}}{2}\right)$$

Aromātai

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