Whakaoti i te -3x^2-11x-10>0

$$- 3 x ^ { 2 } - 11 x - 10 > 0$$

Whakaoti mō x

$$x\in \left(-2,-\frac{5}{3}\right)$$

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 -3x^{2}-11x-10. I te mea he tōraro a -1, ka huri te ahunga koreōrite.

$$3x^{2}+11x+10<0$$

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.

$$3x^{2}+11x+10=0$$

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 3 mō te a, te 11 mō te b, me te 10 mō te c i te ture pūrua.

$$x=\frac{-11±\sqrt{11^{2}-4\times 3\times 10}}{2\times 3}$$

Mahia ngā tātaitai.

$$x=\frac{-11±1}{6}$$

Whakaotia te whārite x=\frac{-11±1}{6} ina he tōrunga te ±, ina he tōraro te ±.

$$x=-\frac{5}{3} x=-2$$

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

$$3\left(x+\frac{5}{3}\right)\left(x+2\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x+\frac{5}{3} me te x+2. Whakaarohia te tauira ina he tōrunga te x+\frac{5}{3} he tōraro te x+2.

$$x+\frac{5}{3}>0 x+2<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x+2 he tōraro te x+\frac{5}{3}.

$$x+2>0 x+\frac{5}{3}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(-2,-\frac{5}{3}\right).

$$x\in \left(-2,-\frac{5}{3}\right)$$

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

$$x\in \left(-2,-\frac{5}{3}\right)$$