Whakaoti i te =-3x^2+11x

$$= - 3 x ^ { 2 } + 11 x$$

Tauwehe

$$x\left(11-3x\right)$$

Ngā Upane Otinga

Tauwehea te x.

$$x\left(-3x+11\right)$$

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.

$$-3x^{2}+11x=0$$

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.

$$x=\frac{-11±\sqrt{11^{2}}}{2\left(-3\right)}$$

Tuhia te pūtakerua o te 11^{2}.

$$x=\frac{-11±11}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$x=\frac{-11±11}{-6}$$

Nā, me whakaoti te whārite x=\frac{-11±11}{-6} ina he tāpiri te ±. Tāpiri -11 ki te 11.

$$x=\frac{0}{-6}$$

Whakawehe 0 ki te -6.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-11±11}{-6} ina he tango te ±. Tango 11 mai i -11.

$$x=-\frac{22}{-6}$$

Whakahekea te hautanga \frac{-22}{-6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{11}{3}$$

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{11}{3} mō te x_{2}.

$$-3x^{2}+11x=-3x\left(x-\frac{11}{3}\right)$$

Tango \frac{11}{3} 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.

$$-3x^{2}+11x=-3x\times \frac{-3x+11}{-3}$$

Whakakorea atu te tauwehe pūnoa nui rawa 3 i roto i te -3 me te -3.

$$-3x^{2}+11x=x\left(-3x+11\right)$$

Aromātai

$$x\left(11-3x\right)$$