Whakaoti i te -2x^2-5x+1

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

Tauwehe

$$-2\left(x-\frac{-\sqrt{33}-5}{4}\right)\left(x-\frac{\sqrt{33}-5}{4}\right)$$

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.

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

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-2\right)}}{2\left(-2\right)}$$

Whakareatia -4 ki te -2.

$$x=\frac{-\left(-5\right)±\sqrt{25+8}}{2\left(-2\right)}$$

Tāpiri 25 ki te 8.

$$x=\frac{-\left(-5\right)±\sqrt{33}}{2\left(-2\right)}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±\sqrt{33}}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$x=\frac{5±\sqrt{33}}{-4}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{33}}{-4} ina he tāpiri te ±. Tāpiri 5 ki te \sqrt{33}.

$$x=\frac{\sqrt{33}+5}{-4}$$

Whakawehe 5+\sqrt{33} ki te -4.

$$x=\frac{-\sqrt{33}-5}{4}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{33}}{-4} ina he tango te ±. Tango \sqrt{33} mai i 5.

$$x=\frac{5-\sqrt{33}}{-4}$$

Whakawehe 5-\sqrt{33} ki te -4.

$$x=\frac{\sqrt{33}-5}{4}$$

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{33}}{4} mō te x_{1} me te \frac{-5+\sqrt{33}}{4} mō te x_{2}.

$$-2x^{2}-5x+1=-2\left(x-\frac{-\sqrt{33}-5}{4}\right)\left(x-\frac{\sqrt{33}-5}{4}\right)$$

Aromātai

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