Whakaoti i te -5x^2+9x+2>0

$$- 5 x ^ { 2 } + 9 x + 2 > 0$$

Whakaoti mō x

$$x\in \left(-\frac{1}{5},2\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me whakarea te koreōrite ki te -1 kia tōrunga ai te tau whakarea o te pū tino teitei i -5x^{2}+9x+2. I te mea he tōraro a -1, ka huri te ahunga koreōrite.

$$5x^{2}-9x-2<0$$

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.

$$5x^{2}-9x-2=0$$

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 5 mō te a, te -9 mō te b, me te -2 mō te c i te ture pūrua.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 5\left(-2\right)}}{2\times 5}$$

Mahia ngā tātaitai.

$$x=\frac{9±11}{10}$$

Whakaotia te whārite x=\frac{9±11}{10} ina he tōrunga te ±, ina he tōraro te ±.

$$x=2 x=-\frac{1}{5}$$

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

$$5\left(x-2\right)\left(x+\frac{1}{5}\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-2 me te x+\frac{1}{5}. Whakaarohia te tauira ina he tōrunga te x-2 he tōraro te x+\frac{1}{5}.

$$x-2>0 x+\frac{1}{5}<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x+\frac{1}{5} he tōraro te x-2.

$$x+\frac{1}{5}>0 x-2<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(-\frac{1}{5},2\right).

$$x\in \left(-\frac{1}{5},2\right)$$

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

$$x\in \left(-\frac{1}{5},2\right)$$