Whakaoti i te -(x+2)(x-5)<0

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

Whakaoti mō x

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Hei kimi i te tauaro o x+2, kimihia te tauaro o ia taurangi.

$$\left(-x-2\right)\left(x-5\right)<0$$

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o -x-2 ki ia tau o x-5.

$$-x^{2}+5x-2x+10<0$$

Pahekotia te 5x me -2x, ka 3x.

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

Me whakarea te koreōrite ki te -1 kia tōrunga ai te tau whakarea o te pū tino teitei i -x^{2}+3x+10. I te mea he tōraro a -1, ka huri te ahunga koreōrite.

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

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.

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

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.

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

Mahia ngā tātaitai.

$$x=\frac{3±7}{2}$$

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

$$x=5 x=-2$$

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

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

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

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

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

$$x<-2$$

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

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

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

$$x>5$$

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

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