Whakaoti i te 12x^2-x-6>0

$$12 x ^ { 2 } - x - 6 > 0$$

Whakaoti mō x

$$x\in \left(-\infty,-\frac{2}{3}\right)\cup \left(\frac{3}{4},\infty\right)$$

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.

$$12x^{2}-x-6=0$$

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 12 mō te a, te -1 mō te b, me te -6 mō te c i te ture pūrua.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 12\left(-6\right)}}{2\times 12}$$

Mahia ngā tātaitai.

$$x=\frac{1±17}{24}$$

Whakaotia te whārite x=\frac{1±17}{24} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{3}{4} x=-\frac{2}{3}$$

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

$$12\left(x-\frac{3}{4}\right)\left(x+\frac{2}{3}\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x-\frac{3}{4} me te x+\frac{2}{3}, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x-\frac{3}{4} me te x+\frac{2}{3}.

$$x-\frac{3}{4}<0 x+\frac{2}{3}<0$$

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

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

Whakaarohia te tauira ina he tōrunga tahi te x-\frac{3}{4} me te x+\frac{2}{3}.

$$x+\frac{2}{3}>0 x-\frac{3}{4}>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>\frac{3}{4}.

$$x>\frac{3}{4}$$

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

$$x<-\frac{2}{3}\text{; }x>\frac{3}{4}$$