Whakaoti i te 12x^2-12^2x+9>0

$$12 { x }^{ 2 } - { 12 }^{ 2 } x+9 > 0$$

Whakaoti mō x

$$x\in \left(-\infty,-\frac{\sqrt{141}}{2}+6\right)\cup \left(\frac{\sqrt{141}}{2}+6,\infty\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tātaihia te 12 mā te pū o 2, kia riro ko 144.

$$12x^{2}-144x+9>0$$

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.

$$12x^{2}-144x+9=0$$

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 12 mō te a, te -144 mō te b, me te 9 mō te c i te ture pūrua.

$$x=\frac{-\left(-144\right)±\sqrt{\left(-144\right)^{2}-4\times 12\times 9}}{2\times 12}$$

Mahia ngā tātaitai.

$$x=\frac{144±12\sqrt{141}}{24}$$

Whakaotia te whārite x=\frac{144±12\sqrt{141}}{24} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{\sqrt{141}}{2}+6 x=-\frac{\sqrt{141}}{2}+6$$

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

$$12\left(x-\left(\frac{\sqrt{141}}{2}+6\right)\right)\left(x-\left(-\frac{\sqrt{141}}{2}+6\right)\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x-\left(\frac{\sqrt{141}}{2}+6\right) me te x-\left(-\frac{\sqrt{141}}{2}+6\right), me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x-\left(\frac{\sqrt{141}}{2}+6\right) me te x-\left(-\frac{\sqrt{141}}{2}+6\right).

$$x-\left(\frac{\sqrt{141}}{2}+6\right)<0 x-\left(-\frac{\sqrt{141}}{2}+6\right)<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<-\frac{\sqrt{141}}{2}+6.

$$x<-\frac{\sqrt{141}}{2}+6$$

Whakaarohia te tauira ina he tōrunga tahi te x-\left(\frac{\sqrt{141}}{2}+6\right) me te x-\left(-\frac{\sqrt{141}}{2}+6\right).

$$x-\left(-\frac{\sqrt{141}}{2}+6\right)>0 x-\left(\frac{\sqrt{141}}{2}+6\right)>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>\frac{\sqrt{141}}{2}+6.

$$x>\frac{\sqrt{141}}{2}+6$$

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

$$x<-\frac{\sqrt{141}}{2}+6\text{; }x>\frac{\sqrt{141}}{2}+6$$