Whakaoti i te 2a^2+a

$$2 a ^ { 2 } + a$$

Tauwehe

$$a\left(2a+1\right)$$

Ngā Upane Otinga

Tauwehea te a.

$$a\left(2a+1\right)$$

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.

$$2a^{2}+a=0$$

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.

$$a=\frac{-1±\sqrt{1^{2}}}{2\times 2}$$

Tuhia te pūtakerua o te 1^{2}.

$$a=\frac{-1±1}{2\times 2}$$

Whakareatia 2 ki te 2.

$$a=\frac{-1±1}{4}$$

Nā, me whakaoti te whārite a=\frac{-1±1}{4} ina he tāpiri te ±. Tāpiri -1 ki te 1.

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

Whakawehe 0 ki te 4.

$$a=0$$

Nā, me whakaoti te whārite a=\frac{-1±1}{4} ina he tango te ±. Tango 1 mai i -1.

$$a=-\frac{2}{4}$$

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

$$a=-\frac{1}{2}$$

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

$$2a^{2}+a=2a\left(a-\left(-\frac{1}{2}\right)\right)$$

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

$$2a^{2}+a=2a\left(a+\frac{1}{2}\right)$$

Tāpiri \frac{1}{2} ki te a 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.

$$2a^{2}+a=2a\times \frac{2a+1}{2}$$

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

$$2a^{2}+a=a\left(2a+1\right)$$

Aromātai

$$a\left(2a+1\right)$$