Whakaoti i te 2x^2+3x+1>0

$$2 { x }^{ 2 } +3x+1 > 0$$

Whakaoti mō x

$$x\in \left(-\infty,-1\right)\cup \left(-\frac{1}{2},\infty\right)$$

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.

$$2x^{2}+3x+1=0$$

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 2 mō te a, te 3 mō te b, me te 1 mō te c i te ture pūrua.

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

Mahia ngā tātaitai.

$$x=\frac{-3±1}{4}$$

Whakaotia te whārite x=\frac{-3±1}{4} ina he tōrunga te ±, ina he tōraro te ±.

$$x=-\frac{1}{2} x=-1$$

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

$$2\left(x+\frac{1}{2}\right)\left(x+1\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x+\frac{1}{2} me te x+1, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x+\frac{1}{2} me te x+1.

$$x+\frac{1}{2}<0 x+1<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<-1.

$$x<-1$$

Whakaarohia te tauira ina he tōrunga tahi te x+\frac{1}{2} me te x+1.

$$x+1>0 x+\frac{1}{2}>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>-\frac{1}{2}.

$$x>-\frac{1}{2}$$

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

$$x<-1\text{; }x>-\frac{1}{2}$$