Whakaoti i te x^2-6x+8>0

$${ x }^{ 2 } -6x+8 > 0$$

Whakaoti mō x

$$x\in \left(-\infty,2\right)\cup \left(4,\infty\right)$$

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.

$$x^{2}-6x+8=0$$

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 8 mō te c i te ture pūrua.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 8}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{6±2}{2}$$

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

$$x=4 x=2$$

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

$$\left(x-4\right)\left(x-2\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x-4 me te x-2, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x-4 me te x-2.

$$x-4<0 x-2<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<2.

$$x<2$$

Whakaarohia te tauira ina he tōrunga tahi te x-4 me te x-2.

$$x-2>0 x-4>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>4.

$$x>4$$

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

$$x<2\text{; }x>4$$