Whakaoti i te 8x^2+25x

$$8 x ^ { 2 } + 25 x$$

Tauwehe

$$x\left(8x+25\right)$$

Ngā Upane Otinga

Tauwehea te x.

$$x\left(8x+25\right)$$

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.

$$8x^{2}+25x=0$$

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.

$$x=\frac{-25±\sqrt{25^{2}}}{2\times 8}$$

Tuhia te pūtakerua o te 25^{2}.

$$x=\frac{-25±25}{2\times 8}$$

Whakareatia 2 ki te 8.

$$x=\frac{-25±25}{16}$$

Nā, me whakaoti te whārite x=\frac{-25±25}{16} ina he tāpiri te ±. Tāpiri -25 ki te 25.

$$x=\frac{0}{16}$$

Whakawehe 0 ki te 16.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-25±25}{16} ina he tango te ±. Tango 25 mai i -25.

$$x=-\frac{50}{16}$$

Whakahekea te hautanga \frac{-50}{16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{25}{8}$$

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{25}{8} mō te x_{2}.

$$8x^{2}+25x=8x\left(x-\left(-\frac{25}{8}\right)\right)$$

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

$$8x^{2}+25x=8x\left(x+\frac{25}{8}\right)$$

Tāpiri \frac{25}{8} 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.

$$8x^{2}+25x=8x\times \frac{8x+25}{8}$$

Whakakorea atu te tauwehe pūnoa nui rawa 8 i roto i te 8 me te 8.

$$8x^{2}+25x=x\left(8x+25\right)$$

Aromātai

$$x\left(8x+25\right)$$