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

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

Tauwehe

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

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.

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

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.

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

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\times 2}}{2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-4±\sqrt{16-8}}{2}$$

Tāpiri 16 ki te -8.

$$x=\frac{-4±\sqrt{8}}{2}$$

Tuhia te pūtakerua o te 8.

$$x=\frac{-4±2\sqrt{2}}{2}$$

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

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

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

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

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

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

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

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

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

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

Aromātai

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