Whakaoti i te f(x)=8x^2-5x

$$f ( x ) = 8 x ^ { 2 } - 5 x$$

Tauwehe

$$x\left(8x-5\right)$$

Ngā Upane Otinga

Tauwehea te x.

$$x\left(8x-5\right)$$

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.

$$8x^{2}-5x=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}}}{2\times 8}$$

Tuhia te pūtakerua o te \left(-5\right)^{2}.

$$x=\frac{-\left(-5\right)±5}{2\times 8}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±5}{2\times 8}$$

Whakareatia 2 ki te 8.

$$x=\frac{5±5}{16}$$

Nā, me whakaoti te whārite x=\frac{5±5}{16} ina he tāpiri te ±. Tāpiri 5 ki te 5.

$$x=\frac{10}{16}$$

Whakahekea te hautanga \frac{10}{16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{5}{8}$$

Nā, me whakaoti te whārite x=\frac{5±5}{16} ina he tango te ±. Tango 5 mai i 5.

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

Whakawehe 0 ki te 16.

$$x=0$$

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{5}{8} mō te x_{1} me te 0 mō te x_{2}.

$$8x^{2}-5x=8\left(x-\frac{5}{8}\right)x$$

Tango \frac{5}{8} mai i x mā te kimi i te tauraro pātahi me te tango i ngā taurunga, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$8x^{2}-5x=8\times \frac{8x-5}{8}x$$

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

$$8x^{2}-5x=\left(8x-5\right)x$$

Aromātai

$$x\left(8x-5\right)$$