Whakaoti i te (x^2+13x+32)

$$( x ^ { 2 } + 13 x + 32 )$$

Tauwehe

$$\left(x-\frac{-\sqrt{41}-13}{2}\right)\left(x-\frac{\sqrt{41}-13}{2}\right)$$

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.

$$x^{2}+13x+32=0$$

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.

$$x=\frac{-13±\sqrt{13^{2}-4\times 32}}{2}$$

Pūrua 13.

$$x=\frac{-13±\sqrt{169-4\times 32}}{2}$$

Whakareatia -4 ki te 32.

$$x=\frac{-13±\sqrt{169-128}}{2}$$

Tāpiri 169 ki te -128.

$$x=\frac{-13±\sqrt{41}}{2}$$

Nā, me whakaoti te whārite x=\frac{-13±\sqrt{41}}{2} ina he tāpiri te ±. Tāpiri -13 ki te \sqrt{41}.

$$x=\frac{\sqrt{41}-13}{2}$$

Nā, me whakaoti te whārite x=\frac{-13±\sqrt{41}}{2} ina he tango te ±. Tango \sqrt{41} mai i -13.

$$x=\frac{-\sqrt{41}-13}{2}$$

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{-13+\sqrt{41}}{2} mō te x_{1} me te \frac{-13-\sqrt{41}}{2} mō te x_{2}.

$$x^{2}+13x+32=\left(x-\frac{\sqrt{41}-13}{2}\right)\left(x-\frac{-\sqrt{41}-13}{2}\right)$$

Aromātai

$$x^{2}+13x+32$$