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

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

Tauwehe

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

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.

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

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\left(-1\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\left(-1\right)}}{2\times 2}$$

Whakareatia -8 ki te -1.

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

Tāpiri 25 ki te 8.

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

Ko te tauaro o -5 ko 5.

$$x=\frac{5±\sqrt{33}}{2\times 2}$$

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}.

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

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

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

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}.

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

Aromātai

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