Whakaoti i te 62x^2+3x-1<0

$$62 x ^ { 2 } + 3 x - 1 < 0$$

Whakaoti mō x

$$x\in \left(\frac{-\sqrt{257}-3}{124},\frac{\sqrt{257}-3}{124}\right)$$

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.

$$62x^{2}+3x-1=0$$

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 62 mō te a, te 3 mō te b, me te -1 mō te c i te ture pūrua.

$$x=\frac{-3±\sqrt{3^{2}-4\times 62\left(-1\right)}}{2\times 62}$$

Mahia ngā tātaitai.

$$x=\frac{-3±\sqrt{257}}{124}$$

Whakaotia te whārite x=\frac{-3±\sqrt{257}}{124} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{\sqrt{257}-3}{124} x=\frac{-\sqrt{257}-3}{124}$$

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

$$62\left(x-\frac{\sqrt{257}-3}{124}\right)\left(x-\frac{-\sqrt{257}-3}{124}\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-\frac{\sqrt{257}-3}{124} me te x-\frac{-\sqrt{257}-3}{124}. Whakaarohia te tauira ina he tōrunga te x-\frac{\sqrt{257}-3}{124} he tōraro te x-\frac{-\sqrt{257}-3}{124}.

$$x-\frac{\sqrt{257}-3}{124}>0 x-\frac{-\sqrt{257}-3}{124}<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x-\frac{-\sqrt{257}-3}{124} he tōraro te x-\frac{\sqrt{257}-3}{124}.

$$x-\frac{-\sqrt{257}-3}{124}>0 x-\frac{\sqrt{257}-3}{124}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(\frac{-\sqrt{257}-3}{124},\frac{\sqrt{257}-3}{124}\right).

$$x\in \left(\frac{-\sqrt{257}-3}{124},\frac{\sqrt{257}-3}{124}\right)$$

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

$$x\in \left(\frac{-\sqrt{257}-3}{124},\frac{\sqrt{257}-3}{124}\right)$$