Whakaoti i te 11x^2-9x+1>0

$$11 { x }^{ 2 } -9x+1 > 0$$

Whakaoti mō x

$$x\in \left(-\infty,\frac{9-\sqrt{37}}{22}\right)\cup \left(\frac{\sqrt{37}+9}{22},\infty\right)$$

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.

$$11x^{2}-9x+1=0$$

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 11 mō te a, te -9 mō te b, me te 1 mō te c i te ture pūrua.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 11\times 1}}{2\times 11}$$

Mahia ngā tātaitai.

$$x=\frac{9±\sqrt{37}}{22}$$

Whakaotia te whārite x=\frac{9±\sqrt{37}}{22} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{\sqrt{37}+9}{22} x=\frac{9-\sqrt{37}}{22}$$

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

$$11\left(x-\frac{\sqrt{37}+9}{22}\right)\left(x-\frac{9-\sqrt{37}}{22}\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x-\frac{\sqrt{37}+9}{22} me te x-\frac{9-\sqrt{37}}{22}, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x-\frac{\sqrt{37}+9}{22} me te x-\frac{9-\sqrt{37}}{22}.

$$x-\frac{\sqrt{37}+9}{22}<0 x-\frac{9-\sqrt{37}}{22}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<\frac{9-\sqrt{37}}{22}.

$$x<\frac{9-\sqrt{37}}{22}$$

Whakaarohia te tauira ina he tōrunga tahi te x-\frac{\sqrt{37}+9}{22} me te x-\frac{9-\sqrt{37}}{22}.

$$x-\frac{9-\sqrt{37}}{22}>0 x-\frac{\sqrt{37}+9}{22}>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>\frac{\sqrt{37}+9}{22}.

$$x>\frac{\sqrt{37}+9}{22}$$

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

$$x<\frac{9-\sqrt{37}}{22}\text{; }x>\frac{\sqrt{37}+9}{22}$$