Whakaoti i te -2x^2+4x+3

$$- 2 x ^ { 2 } + 4 x + 3$$

Tauwehe

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

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.

$$-2x^{2}+4x+3=0$$

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.

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

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\left(-2\right)\times 3}}{2\left(-2\right)}$$

Whakareatia -4 ki te -2.

$$x=\frac{-4±\sqrt{16+8\times 3}}{2\left(-2\right)}$$

Whakareatia 8 ki te 3.

$$x=\frac{-4±\sqrt{16+24}}{2\left(-2\right)}$$

Tāpiri 16 ki te 24.

$$x=\frac{-4±\sqrt{40}}{2\left(-2\right)}$$

Tuhia te pūtakerua o te 40.

$$x=\frac{-4±2\sqrt{10}}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$x=\frac{-4±2\sqrt{10}}{-4}$$

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

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

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

$$x=-\frac{\sqrt{10}}{2}+1$$

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

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

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

$$x=\frac{\sqrt{10}}{2}+1$$

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 1-\frac{\sqrt{10}}{2} mō te x_{1} me te 1+\frac{\sqrt{10}}{2} mō te x_{2}.

$$-2x^{2}+4x+3=-2\left(x-\left(-\frac{\sqrt{10}}{2}+1\right)\right)\left(x-\left(\frac{\sqrt{10}}{2}+1\right)\right)$$

Aromātai

$$3+4x-2x^{2}$$