Whakaoti i te x^2+3x-10>=0

$${ x }^{ 2 } +3x-10 > =0$$

Whakaoti mō x

$$x\in \left(-\infty,-5\right)\cup \left(2,\infty\right)$$

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.

$$x^{2}+3x-10=0$$

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 3 mō te b, me te -10 mō te c i te ture pūrua.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\left(-10\right)}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-3±7}{2}$$

Whakaotia te whārite x=\frac{-3±7}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=2 x=-5$$

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

$$\left(x-2\right)\left(x+5\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x-2 me te x+5, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x-2 me te x+5.

$$x-2<0 x+5<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<-5.

$$x<-5$$

Whakaarohia te tauira ina he tōrunga tahi te x-2 me te x+5.

$$x+5>0 x-2>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>2.

$$x>2$$

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

$$x<-5\text{; }x>2$$