Whakaoti i te x^2+x-4.

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

Tauwehe

$$\left(x-\frac{-\sqrt{17}-1}{2}\right)\left(x-\frac{\sqrt{17}-1}{2}\right)$$

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.

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

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.

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

Pūrua 1.

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

Whakareatia -4 ki te -4.

$$x=\frac{-1±\sqrt{1+16}}{2}$$

Tāpiri 1 ki te 16.

$$x=\frac{-1±\sqrt{17}}{2}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{17}}{2} ina he tāpiri te ±. Tāpiri -1 ki te \sqrt{17}.

$$x=\frac{\sqrt{17}-1}{2}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{17}}{2} ina he tango te ±. Tango \sqrt{17} mai i -1.

$$x=\frac{-\sqrt{17}-1}{2}$$

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 \frac{-1+\sqrt{17}}{2} mō te x_{1} me te \frac{-1-\sqrt{17}}{2} mō te x_{2}.

$$x^{2}+x-4=\left(x-\frac{\sqrt{17}-1}{2}\right)\left(x-\frac{-\sqrt{17}-1}{2}\right)$$

Aromātai

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