Whakaoti i te f(x)=2x^2+10x+1

$$f ( x ) = 2 x ^ { 2 } + 10 x + 1$$

Tauwehe

$$2\left(x-\frac{-\sqrt{23}-5}{2}\right)\left(x-\frac{\sqrt{23}-5}{2}\right)$$

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.

$$2x^{2}+10x+1=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 2}}{2\times 2}$$

Pūrua 10.

$$x=\frac{-10±\sqrt{100-4\times 2}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-10±\sqrt{100-8}}{2\times 2}$$

Tāpiri 100 ki te -8.

$$x=\frac{-10±\sqrt{92}}{2\times 2}$$

Tuhia te pūtakerua o te 92.

$$x=\frac{-10±2\sqrt{23}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-10±2\sqrt{23}}{4}$$

Nā, me whakaoti te whārite x=\frac{-10±2\sqrt{23}}{4} ina he tāpiri te ±. Tāpiri -10 ki te 2\sqrt{23}.

$$x=\frac{2\sqrt{23}-10}{4}$$

Whakawehe -10+2\sqrt{23} ki te 4.

$$x=\frac{\sqrt{23}-5}{2}$$

Nā, me whakaoti te whārite x=\frac{-10±2\sqrt{23}}{4} ina he tango te ±. Tango 2\sqrt{23} mai i -10.

$$x=\frac{-2\sqrt{23}-10}{4}$$

Whakawehe -10-2\sqrt{23} ki te 4.

$$x=\frac{-\sqrt{23}-5}{2}$$

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{23}}{2} mō te x_{1} me te \frac{-5-\sqrt{23}}{2} mō te x_{2}.

$$2x^{2}+10x+1=2\left(x-\frac{\sqrt{23}-5}{2}\right)\left(x-\frac{-\sqrt{23}-5}{2}\right)$$

Aromātai

$$2x^{2}+10x+1$$