Whakaoti i te -2x^2+12x-14>0

$$- 2 x ^ { 2 } + 12 x - 14 > 0$$

Whakaoti mō x

$$x\in \left(3-\sqrt{2},\sqrt{2}+3\right)$$

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 -2x^{2}+12x-14. I te mea he tōraro a -1, ka huri te ahunga koreōrite.

$$2x^{2}-12x+14<0$$

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.

$$2x^{2}-12x+14=0$$

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 -12 mō te b, me te 14 mō te c i te ture pūrua.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\times 14}}{2\times 2}$$

Mahia ngā tātaitai.

$$x=\frac{12±4\sqrt{2}}{4}$$

Whakaotia te whārite x=\frac{12±4\sqrt{2}}{4} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\sqrt{2}+3 x=3-\sqrt{2}$$

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

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

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-\left(\sqrt{2}+3\right) me te x-\left(3-\sqrt{2}\right). Whakaarohia te tauira ina he tōrunga te x-\left(\sqrt{2}+3\right) he tōraro te x-\left(3-\sqrt{2}\right).

$$x-\left(\sqrt{2}+3\right)>0 x-\left(3-\sqrt{2}\right)<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x-\left(3-\sqrt{2}\right) he tōraro te x-\left(\sqrt{2}+3\right).

$$x-\left(3-\sqrt{2}\right)>0 x-\left(\sqrt{2}+3\right)<0$$

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

$$x\in \left(3-\sqrt{2},\sqrt{2}+3\right)$$

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

$$x\in \left(3-\sqrt{2},\sqrt{2}+3\right)$$