Whakaoti i te x^2-5x+6<0

$${ x }^{ 2 } -5x+6 < 0$$

Whakaoti mō x

$$x\in \left(2,3\right)$$

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.

$$x^{2}-5x+6=0$$

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 -5 mō te b, me te 6 mō te c i te ture pūrua.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 6}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{5±1}{2}$$

Whakaotia te whārite x=\frac{5±1}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=3 x=2$$

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

$$\left(x-3\right)\left(x-2\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-3 me te x-2. Whakaarohia te tauira ina he tōrunga te x-3 he tōraro te x-2.

$$x-3>0 x-2<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x-2 he tōraro te x-3.

$$x-2>0 x-3<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(2,3\right).

$$x\in \left(2,3\right)$$

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

$$x\in \left(2,3\right)$$