Whakaoti i te 2x^2+x-1<0

$$2 x ^ { 2 } + x - 1 < 0$$

Whakaoti mō x

$$x\in \left(-1,\frac{1}{2}\right)$$

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.

$$2x^{2}+x-1=0$$

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

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

Mahia ngā tātaitai.

$$x=\frac{-1±3}{4}$$

Whakaotia te whārite x=\frac{-1±3}{4} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{1}{2} x=-1$$

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

$$2\left(x-\frac{1}{2}\right)\left(x+1\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-\frac{1}{2} me te x+1. Whakaarohia te tauira ina he tōrunga te x-\frac{1}{2} he tōraro te x+1.

$$x-\frac{1}{2}>0 x+1<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x+1 he tōraro te x-\frac{1}{2}.

$$x+1>0 x-\frac{1}{2}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(-1,\frac{1}{2}\right).

$$x\in \left(-1,\frac{1}{2}\right)$$

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

$$x\in \left(-1,\frac{1}{2}\right)$$