Whakaoti i te 4x^2+x-8

$$4 x ^ { 2 } + x - 8$$

Tauwehe

$$4\left(x-\frac{-\sqrt{129}-1}{8}\right)\left(x-\frac{\sqrt{129}-1}{8}\right)$$

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.

$$4x^{2}+x-8=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

Pūrua 1.

$$x=\frac{-1±\sqrt{1-4\times 4\left(-8\right)}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-1±\sqrt{1-16\left(-8\right)}}{2\times 4}$$

Whakareatia -16 ki te -8.

$$x=\frac{-1±\sqrt{1+128}}{2\times 4}$$

Tāpiri 1 ki te 128.

$$x=\frac{-1±\sqrt{129}}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-1±\sqrt{129}}{8}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{129}}{8} ina he tāpiri te ±. Tāpiri -1 ki te \sqrt{129}.

$$x=\frac{\sqrt{129}-1}{8}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{129}}{8} ina he tango te ±. Tango \sqrt{129} mai i -1.

$$x=\frac{-\sqrt{129}-1}{8}$$

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{129}}{8} mō te x_{1} me te \frac{-1-\sqrt{129}}{8} mō te x_{2}.

$$4x^{2}+x-8=4\left(x-\frac{\sqrt{129}-1}{8}\right)\left(x-\frac{-\sqrt{129}-1}{8}\right)$$

Aromātai

$$4x^{2}+x-8$$