Whakaoti i te {l}{x^2-x}{-18>0}

$$\left. \begin{array} { l } { x ^ { 2 } - x } \\ { - 18 > 0 } \end{array} \right.$$

Whakaoti mō x

$$x\in \left(-\infty,\frac{1-\sqrt{73}}{2}\right)\cup \left(\frac{\sqrt{73}+1}{2},\infty\right)$$

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.

$$x^{2}-x-18=0$$

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 -1 mō te b, me te -18 mō te c i te ture pūrua.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\left(-18\right)}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{1±\sqrt{73}}{2}$$

Whakaotia te whārite x=\frac{1±\sqrt{73}}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{\sqrt{73}+1}{2} x=\frac{1-\sqrt{73}}{2}$$

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

$$\left(x-\frac{\sqrt{73}+1}{2}\right)\left(x-\frac{1-\sqrt{73}}{2}\right)>0$$

Kia tōrunga te otinga, me tōraro tahi te x-\frac{\sqrt{73}+1}{2} me te x-\frac{1-\sqrt{73}}{2}, me tōrunga tahi rānei. Whakaarohia te tauira ina he tōraro tahi te x-\frac{\sqrt{73}+1}{2} me te x-\frac{1-\sqrt{73}}{2}.

$$x-\frac{\sqrt{73}+1}{2}<0 x-\frac{1-\sqrt{73}}{2}<0$$

Te otinga e whakaea i ngā koreōrite e rua ko x<\frac{1-\sqrt{73}}{2}.

$$x<\frac{1-\sqrt{73}}{2}$$

Whakaarohia te tauira ina he tōrunga tahi te x-\frac{\sqrt{73}+1}{2} me te x-\frac{1-\sqrt{73}}{2}.

$$x-\frac{1-\sqrt{73}}{2}>0 x-\frac{\sqrt{73}+1}{2}>0$$

Te otinga e whakaea i ngā koreōrite e rua ko x>\frac{\sqrt{73}+1}{2}.

$$x>\frac{\sqrt{73}+1}{2}$$

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

$$x<\frac{1-\sqrt{73}}{2}\text{; }x>\frac{\sqrt{73}+1}{2}$$