Whakaoti i te f(x)=2x^2+5x+1

$$f ( x ) = 2 x ^ { 2 } + 5 x + 1$$

Tauwehe

$$2\left(x-\frac{-\sqrt{17}-5}{4}\right)\left(x-\frac{\sqrt{17}-5}{4}\right)$$

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.

$$2x^{2}+5x+1=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2}}{2\times 2}$$

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4\times 2}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-5±\sqrt{25-8}}{2\times 2}$$

Tāpiri 25 ki te -8.

$$x=\frac{-5±\sqrt{17}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-5±\sqrt{17}}{4}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{17}}{4} ina he tāpiri te ±. Tāpiri -5 ki te \sqrt{17}.

$$x=\frac{\sqrt{17}-5}{4}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{17}}{4} ina he tango te ±. Tango \sqrt{17} mai i -5.

$$x=\frac{-\sqrt{17}-5}{4}$$

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}}{4} mō te x_{1} me te \frac{-5-\sqrt{17}}{4} mō te x_{2}.

$$2x^{2}+5x+1=2\left(x-\frac{\sqrt{17}-5}{4}\right)\left(x-\frac{-\sqrt{17}-5}{4}\right)$$

Aromātai

$$2x^{2}+5x+1$$