Whakaoti i te 4-x^2+3x+2x

$$4 - x ^ { 2 } + 3 x + 2 x$$

Tauwehe

$$-\left(x-\frac{5-\sqrt{41}}{2}\right)\left(x-\frac{\sqrt{41}+5}{2}\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Pahekotia te 3x me 2x, ka 5x.

$$factor(4-x^{2}+5x)$$

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.

$$-x^{2}+5x+4=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$x=\frac{-5±\sqrt{25+4\times 4}}{2\left(-1\right)}$$

Whakareatia 4 ki te 4.

$$x=\frac{-5±\sqrt{25+16}}{2\left(-1\right)}$$

Tāpiri 25 ki te 16.

$$x=\frac{-5±\sqrt{41}}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{-5±\sqrt{41}}{-2}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{41}}{-2} ina he tāpiri te ±. Tāpiri -5 ki te \sqrt{41}.

$$x=\frac{\sqrt{41}-5}{-2}$$

Whakawehe -5+\sqrt{41} ki te -2.

$$x=\frac{5-\sqrt{41}}{2}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{41}}{-2} ina he tango te ±. Tango \sqrt{41} mai i -5.

$$x=\frac{-\sqrt{41}-5}{-2}$$

Whakawehe -5-\sqrt{41} ki te -2.

$$x=\frac{\sqrt{41}+5}{2}$$

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 \frac{5-\sqrt{41}}{2} mō te x_{1} me te \frac{5+\sqrt{41}}{2} mō te x_{2}.

$$-x^{2}+5x+4=-\left(x-\frac{5-\sqrt{41}}{2}\right)\left(x-\frac{\sqrt{41}+5}{2}\right)$$

Aromātai

$$4+5x-x^{2}$$

Ngā Upane Otinga

Pahekotia te 3x me 2x, ka 5x.

$$4-x^{2}+5x$$