Whakaoti i te 3x^2-19x+6<0

$$3 { x }^{ 2 } -19x+6 < 0$$

Whakaoti mō x

$$x\in \left(\frac{1}{3},6\right)$$

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.

$$3x^{2}-19x+6=0$$

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 3 mō te a, te -19 mō te b, me te 6 mō te c i te ture pūrua.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 3\times 6}}{2\times 3}$$

Mahia ngā tātaitai.

$$x=\frac{19±17}{6}$$

Whakaotia te whārite x=\frac{19±17}{6} ina he tōrunga te ±, ina he tōraro te ±.

$$x=6 x=\frac{1}{3}$$

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

$$3\left(x-6\right)\left(x-\frac{1}{3}\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-6 me te x-\frac{1}{3}. Whakaarohia te tauira ina he tōrunga te x-6 he tōraro te x-\frac{1}{3}.

$$x-6>0 x-\frac{1}{3}<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x-\frac{1}{3} he tōraro te x-6.

$$x-\frac{1}{3}>0 x-6<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(\frac{1}{3},6\right).

$$x\in \left(\frac{1}{3},6\right)$$

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

$$x\in \left(\frac{1}{3},6\right)$$