Whakaoti i te x^2/3+4/3x-7<0

$$\frac { x ^ { 2 } } { 3 } + \frac { 4 } { 3 } x - 7 < 0$$

Whakaoti mō x

$$x\in \left(-7,3\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia ngā taha e rua o te whārite ki te 3. I te mea he tōrunga te 3, kāore e huri te ahunga koreōrite.

$$x^{2}+4x-21<0$$

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.

$$x^{2}+4x-21=0$$

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 4 mō te b, me te -21 mō te c i te ture pūrua.

$$x=\frac{-4±\sqrt{4^{2}-4\times 1\left(-21\right)}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-4±10}{2}$$

Whakaotia te whārite x=\frac{-4±10}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=3 x=-7$$

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

$$\left(x-3\right)\left(x+7\right)<0$$

Kia tōraro te otinga, me tauaro rawa ngā tohu o te x-3 me te x+7. Whakaarohia te tauira ina he tōrunga te x-3 he tōraro te x+7.

$$x-3>0 x+7<0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Whakaarohia te tauira ina he tōrunga te x+7 he tōraro te x-3.

$$x+7>0 x-3<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left(-7,3\right).

$$x\in \left(-7,3\right)$$

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

$$x\in \left(-7,3\right)$$