Whakaoti i te f(x)=-2x^2-10x+1

$$f ( x ) = - 2 x ^ { 2 } - 10 x + 1$$

Tauwehe

$$-2\left(x-\frac{-3\sqrt{3}-5}{2}\right)\left(x-\frac{3\sqrt{3}-5}{2}\right)$$

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.

$$-2x^{2}-10x+1=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-2\right)}}{2\left(-2\right)}$$

Pūrua -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-2\right)}}{2\left(-2\right)}$$

Whakareatia -4 ki te -2.

$$x=\frac{-\left(-10\right)±\sqrt{100+8}}{2\left(-2\right)}$$

Tāpiri 100 ki te 8.

$$x=\frac{-\left(-10\right)±\sqrt{108}}{2\left(-2\right)}$$

Tuhia te pūtakerua o te 108.

$$x=\frac{-\left(-10\right)±6\sqrt{3}}{2\left(-2\right)}$$

Ko te tauaro o -10 ko 10.

$$x=\frac{10±6\sqrt{3}}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$x=\frac{10±6\sqrt{3}}{-4}$$

Nā, me whakaoti te whārite x=\frac{10±6\sqrt{3}}{-4} ina he tāpiri te ±. Tāpiri 10 ki te 6\sqrt{3}.

$$x=\frac{6\sqrt{3}+10}{-4}$$

Whakawehe 10+6\sqrt{3} ki te -4.

$$x=\frac{-3\sqrt{3}-5}{2}$$

Nā, me whakaoti te whārite x=\frac{10±6\sqrt{3}}{-4} ina he tango te ±. Tango 6\sqrt{3} mai i 10.

$$x=\frac{10-6\sqrt{3}}{-4}$$

Whakawehe 10-6\sqrt{3} ki te -4.

$$x=\frac{3\sqrt{3}-5}{2}$$

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{-5-3\sqrt{3}}{2} mō te x_{1} me te \frac{-5+3\sqrt{3}}{2} mō te x_{2}.

$$-2x^{2}-10x+1=-2\left(x-\frac{-3\sqrt{3}-5}{2}\right)\left(x-\frac{3\sqrt{3}-5}{2}\right)$$

Aromātai

$$1-10x-2x^{2}$$