Whakaoti i te -3x^2-8x+3<0

$$- 3 x ^ { 2 } - 8 x + 3 < 0$$

Whakaoti mō x

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

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 -3x^{2}-8x+3. I te mea he tōraro a -1, ka huri te ahunga koreōrite.

$$3x^{2}+8x-3>0$$

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.

$$3x^{2}+8x-3=0$$

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 3 mō te a, te 8 mō te b, me te -3 mō te c i te ture pūrua.

$$x=\frac{-8±\sqrt{8^{2}-4\times 3\left(-3\right)}}{2\times 3}$$

Mahia ngā tātaitai.

$$x=\frac{-8±10}{6}$$

Whakaotia te whārite x=\frac{-8±10}{6} ina he tōrunga te ±, ina he tōraro te ±.

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

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

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

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

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

Te otinga e whakaea i ngā koreōrite e rua ko x<-3.

$$x<-3$$

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

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

Te otinga e whakaea i ngā koreōrite e rua ko x>\frac{1}{3}.

$$x>\frac{1}{3}$$

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

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