Whakaoti i te 3x^2+13x-10<0

$$3 x ^ { 2 } + 13 x - 10 < 0$$

Whakaoti mō x

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

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.

$$3x^{2}+13x-10=0$$

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 13 mō te b, me te -10 mō te c i te ture pūrua.

$$x=\frac{-13±\sqrt{13^{2}-4\times 3\left(-10\right)}}{2\times 3}$$

Mahia ngā tātaitai.

$$x=\frac{-13±17}{6}$$

Whakaotia te whārite x=\frac{-13±17}{6} ina he tōrunga te ±, ina he tōraro te ±.

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

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

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

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

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

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

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

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

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

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

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

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