Whakaoti i te -2x^2-5x-3<0

$$- 2 x ^ { 2 } - 5 x - 3 < 0$$

Whakaoti mō x

$$x\in \left(-\infty,-\frac{3}{2}\right)\cup \left(-1,\infty\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me whakarea te koreōrite ki te -1 kia tōrunga ai te tau whakarea o te pū tino teitei i -2x^{2}-5x-3. I te mea he tōraro a -1, ka huri te ahunga koreōrite.

$$2x^{2}+5x+3>0$$

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.

$$2x^{2}+5x+3=0$$

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

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\times 3}}{2\times 2}$$

Mahia ngā tātaitai.

$$x=\frac{-5±1}{4}$$

Whakaotia te whārite x=\frac{-5±1}{4} ina he tōrunga te ±, ina he tōraro te ±.

$$x=-1 x=-\frac{3}{2}$$

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

$$2\left(x+1\right)\left(x+\frac{3}{2}\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x+1 me te x+\frac{3}{2}, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x+1 me te x+\frac{3}{2}.

$$x+1<0 x+\frac{3}{2}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<-\frac{3}{2}.

$$x<-\frac{3}{2}$$

Whakaarohia te tauira ina he tōrunga tahi te x+1 me te x+\frac{3}{2}.

$$x+\frac{3}{2}>0 x+1>0$$

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

$$x>-1$$

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

$$x<-\frac{3}{2}\text{; }x>-1$$