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

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

Tauwehe

$$-2\left(x-\frac{1-\sqrt{41}}{4}\right)\left(x-\frac{\sqrt{41}+1}{4}\right)$$

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.

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

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.

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

Pūrua 1.

$$x=\frac{-1±\sqrt{1-4\left(-2\right)\times 5}}{2\left(-2\right)}$$

Whakareatia -4 ki te -2.

$$x=\frac{-1±\sqrt{1+8\times 5}}{2\left(-2\right)}$$

Whakareatia 8 ki te 5.

$$x=\frac{-1±\sqrt{1+40}}{2\left(-2\right)}$$

Tāpiri 1 ki te 40.

$$x=\frac{-1±\sqrt{41}}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$x=\frac{-1±\sqrt{41}}{-4}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{41}}{-4} ina he tāpiri te ±. Tāpiri -1 ki te \sqrt{41}.

$$x=\frac{\sqrt{41}-1}{-4}$$

Whakawehe -1+\sqrt{41} ki te -4.

$$x=\frac{1-\sqrt{41}}{4}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{41}}{-4} ina he tango te ±. Tango \sqrt{41} mai i -1.

$$x=\frac{-\sqrt{41}-1}{-4}$$

Whakawehe -1-\sqrt{41} ki te -4.

$$x=\frac{\sqrt{41}+1}{4}$$

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{1-\sqrt{41}}{4} mō te x_{1} me te \frac{1+\sqrt{41}}{4} mō te x_{2}.

$$-2x^{2}+x+5=-2\left(x-\frac{1-\sqrt{41}}{4}\right)\left(x-\frac{\sqrt{41}+1}{4}\right)$$

Aromātai

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