Whakaoti i te 4x^2+1x

$$4 x ^ { 2 } + 1 x$$

Tauwehe

$$x\left(4x+1\right)$$

Ngā Upane Otinga

Tauwehea te x.

$$x\left(4x+1\right)$$

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.

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

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.

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

Tuhia te pūtakerua o te 1^{2}.

$$x=\frac{-1±1}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-1±1}{8}$$

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

$$x=\frac{0}{8}$$

Whakawehe 0 ki te 8.

$$x=0$$

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

$$x=-\frac{2}{8}$$

Whakahekea te hautanga \frac{-2}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{1}{4}$$

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}{4} mō te x_{2}.

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

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

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

Tāpiri \frac{1}{4} ki te x 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.

$$4x^{2}+x=4x\times \frac{4x+1}{4}$$

Whakakorea atu te tauwehe pūnoa nui rawa 4 i roto i te 4 me te 4.

$$4x^{2}+x=x\left(4x+1\right)$$

Aromātai

$$x\left(4x+1\right)$$