Whakaoti i te x^2-20x+40

$${ x }^{ 2 } -20x+40$$

Tauwehe

$$\left(x-\left(10-2\sqrt{15}\right)\right)\left(x-\left(2\sqrt{15}+10\right)\right)$$

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.

$$x^{2}-20x+40=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 40}}{2}$$

Pūrua -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 40}}{2}$$

Whakareatia -4 ki te 40.

$$x=\frac{-\left(-20\right)±\sqrt{400-160}}{2}$$

Tāpiri 400 ki te -160.

$$x=\frac{-\left(-20\right)±\sqrt{240}}{2}$$

Tuhia te pūtakerua o te 240.

$$x=\frac{-\left(-20\right)±4\sqrt{15}}{2}$$

Ko te tauaro o -20 ko 20.

$$x=\frac{20±4\sqrt{15}}{2}$$

Nā, me whakaoti te whārite x=\frac{20±4\sqrt{15}}{2} ina he tāpiri te ±. Tāpiri 20 ki te 4\sqrt{15}.

$$x=\frac{4\sqrt{15}+20}{2}$$

Whakawehe 20+4\sqrt{15} ki te 2.

$$x=2\sqrt{15}+10$$

Nā, me whakaoti te whārite x=\frac{20±4\sqrt{15}}{2} ina he tango te ±. Tango 4\sqrt{15} mai i 20.

$$x=\frac{20-4\sqrt{15}}{2}$$

Whakawehe 20-4\sqrt{15} ki te 2.

$$x=10-2\sqrt{15}$$

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 10+2\sqrt{15} mō te x_{1} me te 10-2\sqrt{15} mō te x_{2}.

$$x^{2}-20x+40=\left(x-\left(2\sqrt{15}+10\right)\right)\left(x-\left(10-2\sqrt{15}\right)\right)$$

Aromātai

$$x^{2}-20x+40$$