Whakaoti i te n^2+9n+4

$$n ^ { 2 } + 9 n + 4$$

Tauwehe

$$\left(n-\frac{-\sqrt{65}-9}{2}\right)\left(n-\frac{\sqrt{65}-9}{2}\right)$$

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.

$$n^{2}+9n+4=0$$

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.

$$n=\frac{-9±\sqrt{9^{2}-4\times 4}}{2}$$

Pūrua 9.

$$n=\frac{-9±\sqrt{81-4\times 4}}{2}$$

Whakareatia -4 ki te 4.

$$n=\frac{-9±\sqrt{81-16}}{2}$$

Tāpiri 81 ki te -16.

$$n=\frac{-9±\sqrt{65}}{2}$$

Nā, me whakaoti te whārite n=\frac{-9±\sqrt{65}}{2} ina he tāpiri te ±. Tāpiri -9 ki te \sqrt{65}.

$$n=\frac{\sqrt{65}-9}{2}$$

Nā, me whakaoti te whārite n=\frac{-9±\sqrt{65}}{2} ina he tango te ±. Tango \sqrt{65} mai i -9.

$$n=\frac{-\sqrt{65}-9}{2}$$

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{-9+\sqrt{65}}{2} mō te x_{1} me te \frac{-9-\sqrt{65}}{2} mō te x_{2}.

$$n^{2}+9n+4=\left(n-\frac{\sqrt{65}-9}{2}\right)\left(n-\frac{-\sqrt{65}-9}{2}\right)$$

Aromātai

$$n^{2}+9n+4$$