Whakaoti i te 2f^2+5f

$$2 f ^ { 2 } + 5 f$$

Tauwehe

$$f\left(2f+5\right)$$

Ngā Upane Otinga

Tauwehea te f.

$$f\left(2f+5\right)$$

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.

$$2f^{2}+5f=0$$

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.

$$f=\frac{-5±\sqrt{5^{2}}}{2\times 2}$$

Tuhia te pūtakerua o te 5^{2}.

$$f=\frac{-5±5}{2\times 2}$$

Whakareatia 2 ki te 2.

$$f=\frac{-5±5}{4}$$

Nā, me whakaoti te whārite f=\frac{-5±5}{4} ina he tāpiri te ±. Tāpiri -5 ki te 5.

$$f=\frac{0}{4}$$

Whakawehe 0 ki te 4.

$$f=0$$

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

$$f=-\frac{10}{4}$$

Whakahekea te hautanga \frac{-10}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$f=-\frac{5}{2}$$

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 0 mō te x_{1} me te -\frac{5}{2} mō te x_{2}.

$$2f^{2}+5f=2f\left(f-\left(-\frac{5}{2}\right)\right)$$

Whakamāmātia ngā kīanga katoa o te āhua p-\left(-q\right) ki te p+q.

$$2f^{2}+5f=2f\left(f+\frac{5}{2}\right)$$

Tāpiri \frac{5}{2} ki te f mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$2f^{2}+5f=2f\times \frac{2f+5}{2}$$

Whakakorea atu te tauwehe pūnoa nui rawa 2 i roto i te 2 me te 2.

$$2f^{2}+5f=f\left(2f+5\right)$$

Aromātai

$$f\left(2f+5\right)$$