Whakaoti i te x^2+10x+24>0

$$x ^ { 2 } + 10 x + 24 > 0$$

Whakaoti mō x

$$x\in \left(-\infty,-6\right)\cup \left(-4,\infty\right)$$

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.

$$x^{2}+10x+24=0$$

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 10 mō te b, me te 24 mō te c i te ture pūrua.

$$x=\frac{-10±\sqrt{10^{2}-4\times 1\times 24}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-10±2}{2}$$

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

$$x=-4 x=-6$$

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

$$\left(x+4\right)\left(x+6\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x+4 me te x+6, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x+4 me te x+6.

$$x+4<0 x+6<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<-6.

$$x<-6$$

Whakaarohia te tauira ina he tōrunga tahi te x+4 me te x+6.

$$x+6>0 x+4>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>-4.

$$x>-4$$

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

$$x<-6\text{; }x>-4$$