Whakaoti i te 3x^2+7x-20>0

$$3 x ^ { 2 } + 7 x - 20 > 0$$

Whakaoti mō x

$$x\in \left(-\infty,-4\right)\cup \left(\frac{5}{3},\infty\right)$$

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.

$$3x^{2}+7x-20=0$$

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 7 mō te b, me te -20 mō te c i te ture pūrua.

$$x=\frac{-7±\sqrt{7^{2}-4\times 3\left(-20\right)}}{2\times 3}$$

Mahia ngā tātaitai.

$$x=\frac{-7±17}{6}$$

Whakaotia te whārite x=\frac{-7±17}{6} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{5}{3} x=-4$$

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

$$3\left(x-\frac{5}{3}\right)\left(x+4\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x-\frac{5}{3} me te x+4, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x-\frac{5}{3} me te x+4.

$$x-\frac{5}{3}<0 x+4<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<-4.

$$x<-4$$

Whakaarohia te tauira ina he tōrunga tahi te x-\frac{5}{3} me te x+4.

$$x+4>0 x-\frac{5}{3}>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>\frac{5}{3}.

$$x>\frac{5}{3}$$

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

$$x<-4\text{; }x>\frac{5}{3}$$