Whakaoti i te 2x(x+3)-2(3x+5)+x>0

$$2 x ( x + 3 ) - 2 ( 3 x + 5 ) + x > 0$$

Whakaoti mō x

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 2x ki te x+3.

$$2x^{2}+6x-2\left(3x+5\right)+x>0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -2 ki te 3x+5.

$$2x^{2}+6x-6x-10+x>0$$

Pahekotia te 6x me -6x, ka 0.

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

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.

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

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 2 mō te a, te 1 mō te b, me te -10 mō te c i te ture pūrua.

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

Mahia ngā tātaitai.

$$x=\frac{-1±9}{4}$$

Whakaotia te whārite x=\frac{-1±9}{4} ina he tōrunga te ±, ina he tōraro te ±.

$$x=2 x=-\frac{5}{2}$$

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

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

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

$$x-2<0 x+\frac{5}{2}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<-\frac{5}{2}.

$$x<-\frac{5}{2}$$

Whakaarohia te tauira ina he tōrunga tahi te x-2 me te x+\frac{5}{2}.

$$x+\frac{5}{2}>0 x-2>0$$

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.

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