Whakaoti i te x^2-6x+2>0

$$x ^ { 2 } - 6 x + 2 > 0$$

Whakaoti mō x

$$x\in \left(-\infty,3-\sqrt{7}\right)\cup \left(\sqrt{7}+3,\infty\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Kia whakaotia te koreōrite, me tauwehe te taha mauī. 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.

$$x^{2}-6x+2=0$$

Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te 1 mō te a, te -6 mō te b, me te 2 mō te c i te ture pūrua.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 2}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{6±2\sqrt{7}}{2}$$

Whakaotia te whārite x=\frac{6±2\sqrt{7}}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\sqrt{7}+3 x=3-\sqrt{7}$$

Tuhia anō te koreōrite mā te whakamahi i ngā otinga i whiwhi.

$$\left(x-\left(\sqrt{7}+3\right)\right)\left(x-\left(3-\sqrt{7}\right)\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x-\left(\sqrt{7}+3\right) me te x-\left(3-\sqrt{7}\right), me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x-\left(\sqrt{7}+3\right) me te x-\left(3-\sqrt{7}\right).

$$x-\left(\sqrt{7}+3\right)<0 x-\left(3-\sqrt{7}\right)<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<3-\sqrt{7}.

$$x<3-\sqrt{7}$$

Whakaarohia te tauira ina he tōrunga tahi te x-\left(\sqrt{7}+3\right) me te x-\left(3-\sqrt{7}\right).

$$x-\left(3-\sqrt{7}\right)>0 x-\left(\sqrt{7}+3\right)>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>\sqrt{7}+3.

$$x>\sqrt{7}+3$$

Ko te otinga whakamutunga ko te whakakotahi i ngā otinga kua whiwhi.

$$x<3-\sqrt{7}\text{; }x>\sqrt{7}+3$$