Whakaoti i te 20x^2+x-1>0

$$20 x ^ { 2 } + x - 1 > 0$$

Whakaoti mō x

$$x\in \left(-\infty,-\frac{1}{4}\right)\cup \left(\frac{1}{5},\infty\right)$$

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.

$$20x^{2}+x-1=0$$

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

$$x=\frac{-1±\sqrt{1^{2}-4\times 20\left(-1\right)}}{2\times 20}$$

Mahia ngā tātaitai.

$$x=\frac{-1±9}{40}$$

Whakaotia te whārite x=\frac{-1±9}{40} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{1}{5} x=-\frac{1}{4}$$

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

$$20\left(x-\frac{1}{5}\right)\left(x+\frac{1}{4}\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x-\frac{1}{5} me te x+\frac{1}{4}, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x-\frac{1}{5} me te x+\frac{1}{4}.

$$x-\frac{1}{5}<0 x+\frac{1}{4}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<-\frac{1}{4}.

$$x<-\frac{1}{4}$$

Whakaarohia te tauira ina he tōrunga tahi te x-\frac{1}{5} me te x+\frac{1}{4}.

$$x+\frac{1}{4}>0 x-\frac{1}{5}>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>\frac{1}{5}.

$$x>\frac{1}{5}$$

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

$$x<-\frac{1}{4}\text{; }x>\frac{1}{5}$$