Whakaoti i te -x^2+8>2x

$$- { x }^{ 2 } +8 > 2x$$

Whakaoti mō x

$$x\in \left(-4,2\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 2x mai i ngā taha e rua.

$$-x^{2}+8-2x>0$$

Me whakarea te koreōrite ki te -1 kia tōrunga ai te tau whakarea o te pū tino teitei i -x^{2}+8-2x. I te mea he tōraro a -1, ka huri te ahunga koreōrite.

$$x^{2}-8+2x<0$$

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.

$$x^{2}-8+2x=0$$

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 1 mō te a, te 2 mō te b, me te -8 mō te c i te ture pūrua.

$$x=\frac{-2±\sqrt{2^{2}-4\times 1\left(-8\right)}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-2±6}{2}$$

Whakaotia te whārite x=\frac{-2±6}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=2 x=-4$$

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

$$\left(x-2\right)\left(x+4\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-2 me te x+4. Whakaarohia te tauira ina he tōrunga te x-2 he tōraro te x+4.

$$x-2>0 x+4<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x+4 he tōraro te x-2.

$$x+4>0 x-2<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(-4,2\right).

$$x\in \left(-4,2\right)$$

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

$$x\in \left(-4,2\right)$$