Whakaoti i te x^2+3x+2<0

$$x ^ { 2 } + 3 x + 2 < 0$$

Whakaoti mō x

$$x\in \left(-2,-1\right)$$

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.

$$x^{2}+3x+2=0$$

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 3 mō te b, me te 2 mō te c i te ture pūrua.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\times 2}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-3±1}{2}$$

Whakaotia te whārite x=\frac{-3±1}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=-1 x=-2$$

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

$$\left(x+1\right)\left(x+2\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x+1 me te x+2. Whakaarohia te tauira ina he tōrunga te x+1 he tōraro te x+2.

$$x+1>0 x+2<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x+2 he tōraro te x+1.

$$x+2>0 x+1<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(-2,-1\right).

$$x\in \left(-2,-1\right)$$

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

$$x\in \left(-2,-1\right)$$