Whakaoti i te x^2+9x-3>0

$$x ^ { 2 } + 9 x - 3 > 0$$

Whakaoti mō x

$$x\in \left(-\infty,\frac{-\sqrt{93}-9}{2}\right)\cup \left(\frac{\sqrt{93}-9}{2},\infty\right)$$

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.

$$x^{2}+9x-3=0$$

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

$$x=\frac{-9±\sqrt{9^{2}-4\times 1\left(-3\right)}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-9±\sqrt{93}}{2}$$

Whakaotia te whārite x=\frac{-9±\sqrt{93}}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{\sqrt{93}-9}{2} x=\frac{-\sqrt{93}-9}{2}$$

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

$$\left(x-\frac{\sqrt{93}-9}{2}\right)\left(x-\frac{-\sqrt{93}-9}{2}\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x-\frac{\sqrt{93}-9}{2} me te x-\frac{-\sqrt{93}-9}{2}, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x-\frac{\sqrt{93}-9}{2} me te x-\frac{-\sqrt{93}-9}{2}.

$$x-\frac{\sqrt{93}-9}{2}<0 x-\frac{-\sqrt{93}-9}{2}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<\frac{-\sqrt{93}-9}{2}.

$$x<\frac{-\sqrt{93}-9}{2}$$

Whakaarohia te tauira ina he tōrunga tahi te x-\frac{\sqrt{93}-9}{2} me te x-\frac{-\sqrt{93}-9}{2}.

$$x-\frac{-\sqrt{93}-9}{2}>0 x-\frac{\sqrt{93}-9}{2}>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>\frac{\sqrt{93}-9}{2}.

$$x>\frac{\sqrt{93}-9}{2}$$

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

$$x<\frac{-\sqrt{93}-9}{2}\text{; }x>\frac{\sqrt{93}-9}{2}$$